These extra practice problems for Class 1 come from exams and review exercises that I wrote for Econ 103, from past Oxford Prelims collections, and from past Oxford demonstration lectures. Econ 103 is the introductory statistics course that I taught at the University of Pennsylvania. They are grouped by topic. Within each topic they run from warm-up questions, through standard and harder problems, to challenge problems that go beyond what the lecture videos cover. Try each problem before clicking to reveal its solution.
Conditional probability
Problem 7 Warm-up
State the general multiplication rule, and compare it to the definition of conditional probability.
The multiplication rule states that \(P(A\cap B) = P(A|B)P(B)\). It is the definition of conditional probability, \(P(A|B) = P(A\cap B)/P(B)\), with both sides multiplied by \(P(B)\).
Problem 8 Warm-up
Suppose I flip a fair coin and roll a single fair die at the same time. Define the events
\(A =\) the coin comes up tails
\(B =\) the die shows a 3 or 5
\(C =\) the die shows an odd number
Calculate \(P(B|C)\).
\[P(B|C) = P(B\cap C)/P(C) = (1/3)/(1/2) = 2/3\]
Calculate \(P(A \cap B)\).
Since the dice roll and coin flip are independent, we have \(P(A \cap B) = P(A)P(B) = (1/2) \times (1/3) = 1/6\). You could also draw out the table with all 12 basic outcomes, all of which are equally likely, and count how many are in both \(A\) and \(B\). This will give you \(2/12 = 1/6\)
Calculate \(P(A \cup B)\).
\(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/2 + 1/3 -1/6 = 3/6 + 2/6 - 1/6 = 4/6 = 2/3\). Again, you could also draw out the table with all 12 basic outcomes, all of which are equally likely, and count how many are in either \(A\), \(B\) or both \(A\) and \(B\). This will give you \(8/12 = 2/3\)
Problem 9 Warm-up
Assa made a cross-tab classifying all Penn undergraduates by their class standing in 2017–2018 (freshman, sophomore, junior, or senior) and the number of times that they visited the Penn museum during the academic year (never, once, or more than once).
|
|
Museum Visits |
|
|
Never |
Once |
More than Once |
|
Freshman |
0.08 |
0.10 |
0.04 |
| Class |
Sophomore |
0.04 |
0.10 |
0.04 |
| Junior |
0.04 |
0.20 |
0.09 |
|
Senior |
0.02 |
0.15 |
0.10 |
For example, we see that 8% of Penn undergraduates in 2017–2018 were Freshmen who never visited the Museum, while 10% were seniors who visited more than once.
What fraction of undergraduates visited the museum more than once?
\(P(\text{More than Once}) = 0.04 + 0.04 + 0.09 + 0.10 = 0.27\)
Suppose we chose a random Penn undergraduate in 2017–2018. If she is a junior, what is the probability that she never visited the museum?
Since \(P(\text{Junior}) = 0.04 + 0.20 + 0.09 = 0.33\), \[P(\text{Never}|\text{Junior}) = \frac{P(\text{Never}\cap \text{Junior})}{P(\text{Junior})} = \frac{0.04}{0.33} = \frac{4}{33}\]
Problem 10 Standard
Suppose that I choose two distinct numbers at random from the set \(\left\{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \right\}\). What is the probability that both are odd?
Choose the numbers one at a time. By the multiplication rule, \(P(\text{both odd}) = P(\text{first odd})P(\text{second odd}|\text{first odd}) = (5/9)(4/8) = 5/18\). Alternatively, count pairs, assuming that order doesn’t matter. You could also assume that it does matter and get the same answer. There are \(\displaystyle {9 \choose 2} = 36\) equally likely ways to choose 2 items from a set of 9. Of these, there are \(\displaystyle {5 \choose 2} = 10\) ways to choose 2 of the 5 odd numbers. Hence the probability is \(10/36 = 5/18\).
Independence
Problem 11 Warm-up
Define statistical independence. How is it related to conditional probability, and what does it mean intuitively?
Events \(A\) and \(B\) are independent if \(P(A\cap B) = P(A)P(B)\). Provided \(P(B) > 0\), this is equivalent to \(P(A|B) = P(A)\). Intuitively, learning that \(B\) has occurred tells us nothing about whether \(A\) has occurred.
Problem 12 Standard
Suppose a couple decides to have three children. Assume that the sex of each child is independent, and the probability of a girl is \(0.48\). (Both of these assumptions are approximately correct for US data.) Note that order matters in this example.
How many basic outcomes are there for this experiment?
There are two possible outcomes for each birth, so the total number of possibilities is \(2\times 2\times 2 = 8\).
Are the basic outcomes equally likely? Why or why not?
No, because each child is more likely to be a boy than a girl. The outcome BBB is most likely, followed by outcomes with two boys, and then outcomes with one boy. The outcome GGG is least likely.
What is the probability that the couple has at least one girl?
Use the Complement Rule and independence to calculate the probability of no girls, i.e. all boys: \[0.52 \times 0.52 \times 0.52 \approx 0.14\] Hence, the probability of at least one girl is approximately \(1 - 0.14 = 0.86\)
Problem 13 Standard
What is the probability of getting at least one six when rolling a fair, six-sided die three times? Explain your answer.
Using the complement rule: \[P(\text{At Least One Six}) = 1 - P(\text{No Sixes})\] And by independence: \[P(\text{No Sixes}) = 5/6 \times 5/6 \times 5/6 = 125/216\] Hence, \[P(\text{At Least One Six}) = 1 - 125/216 = 91/216 \approx 0.42\]
Problem 14 Standard
Suppose I throw two fair, six-sided dice once. Define the following events: \[\begin{aligned}
E &= \text{The first die shows 5}\\
F &= \text{The sum of the two dice equals 7}\\
G &= \text{The sum of the two dice equals 10}
\end{aligned}\]
What is \(P(F)\)?
Of the 36 basic outcomes of the experiment, the pairs (1,6), (6,1), (2,5), (5,2), (3,4), and (4,3) sum to 7. Hence the probability is 1/6.
What is \(P(G)\)?
Of the 36 basic outcomes of this experiment, the pairs (5,5), (4,6), and (6,4) sum to 10. Hence the probability is \(3/36 = 1/12\).
Calculate \(P(F|E)\).
By the definition of conditional probability, \[P(F|E) = \frac{P(F\cap E)}{P(E)}\] We know that \(P(E) = 1/6\). The only way that \(F\cap E\) can occur is if we roll (5,2). Hence \(P(F\cap E) = 1/36\). Thus, \(P(F|E) = (1/36)/(1/6) = 6/36 = 1/6\).
Calculate \(P(G|E)\).
Again, by the definition of conditional probability, \[P(G|E) = \frac{P(G\cap E)}{P(E)}\] As before, \(P(E) = 1/6\). The only way for \(G\cap E\) to occur is if we roll (5,5). Hence \(P(G|E) = (1/36)/(1/6) = 6/36 = 1/6\).
Suppose you wanted to bet on the outcome of this random experiment. If I revealed whether or not the first die was a 5, would this give you any relevant information for betting on \(F\)? What about for betting on \(G\)? Explain briefly.
Using the results of the preceding parts, we see that \(P(F|E) =P(F)\), hence \(F\) and \(E\) are statistically independent. This means that knowing whether or not \(E\) has occurred gives us no additional information about the probability of \(F\). In contrast \(P(G|E)\neq P(G)\), so \(G\) and \(E\) are not statistically independent. Indeed, if we know that \(E\) has occurred, this doubles the chance that \(G\) has occurred.
Problem 15 Standard
Suppose I flip a fair coin three times. Let \(A\) be the event that I get at least one head, and \(B\) be the event that I get exactly two heads.
Calculate \(P(B|A)\).
This experiment has \(2 \times 2 \times 2 = 8\) equally likely basic outcomes. Of these, 7 have at least one head:
| HHH |
THH |
HTH |
HHT |
HTT |
THT |
TTH |
Hence, \(P(A) = 7/8\). Among the 7 basic outcomes with at least one head, 3 have exactly two heads, namely:
Thus, \(P(B \cap A) = 3/8\). By the definition of conditional probability, the desired probability is \(P(B|A) = P(B\cap A)/P(A) = (3/8) / (7/8) = 3/7\).
Are the events \(A\) and \(B\) independent? Justify your answer.
Of the 8 equally likely basic outcomes, 3 have exactly two heads, namely HHT, HTH, and THH. Thus, \(P(B) = 3/8\). From the preceding part, however, we know that \(P(B|A) = 3/7\). Since knowing that \(A\) has occurred changes the probability that \(B\) will occur, \(A\) and \(B\) are not independent. Another way to argue this is by pointing out that \(P(A) \times P(B) = \frac{7}{8} \times \frac{3}{8} = \frac{21}{64}\) whereas \(P(A\cap B) = 3/8\).
Problem 16 Harder
You have been entered into a very strange tennis tournament. To get the $10,000 Grand Prize you must win at least two sets in a row in a three-set series to be played against your professor and Venus Williams alternately: professor-Venus-professor or Venus-professor-Venus according to your choice. Let \(p\) be the probability that you win a set against your professor and \(v\) be the probability that you win a set against Venus. Naturally \(p>v\) since Venus is much better than your professor! Assume that each set is independent.
Let W indicate win and L indicate lose, so that the sequence WWW means you win all three sets, WLW means you win the first and third set but lose the middle one, and so on. Which sequences of wins and losses land you the Grand Prize?
To get the prize, you have to win the middle set. Thus, the only possibilities are WWW, WWL, and LWW.
If you elect to play the middle set against Venus, what is the probability that you win the Grand Prize?
The probabilities of mutually exclusive events sum. Thus, \[\begin{aligned}
P(WWW) + P(LWW) + P(WWL) &= pvp + (1-p)vp +pv(1-p)\\
&= p^2v + pv - p^2v + pv - p^2v\\
&= 2pv - p^2v \\
&= pv(2-p)
\end{aligned}\]
If you elect to play the middle set against your professor, what is the probability that you win the Grand prize?
Again, the probabilities of mutually exclusive events sum. Thus, \[\begin{aligned}
P(WWW) + P(LWW) + P(WWL) &= vpv + (1-v)pv +vp(1-v)\\
&=v^2 p + vp - v^2 p + vp - v^2 p\\
&= 2pv -v^2p\\
&= pv(2-v)
\end{aligned}\]
To maximize your chance of winning the prize, should you choose to play the middle set against Venus or your professor?
Manipulating the inequality, \[\begin{aligned}
p &> v\\
-p &< -v\\
2-p &< 2-v\\
pv(2-p) &< pv(2-v)\\
\end{aligned}\] You can’t get the prize without winning the middle set, so it turns out that it’s better to face Venus twice rather than face her in the middle set. You should elect to play the middle set against your professor.
Problem 17 Harder
Rossa and Rodrigo are playing their favorite game: matching pennies. The game proceeds as follows. In each round, each player flips a penny. If the flips match (TT or HH) Rossa gets one point; if the flips do not match (TH or HT) Rodrigo gets one point. The game is best of three rounds: as soon as one of the players reaches two points, the game ends and that player is declared the winner. Since there’s a lot of money on the line and graduate students aren’t paid particularly well, Rossa secretly alters each of the pennies so that the probability of heads is 2/3 rather than 1/2. In spite of Rossa’s cheating, the individual coin flips remain independent.
Calculate the probability that Rossa will win the first round of this game.
Rossa wins a given round if either of the two mutually exclusive outcomes \(HH\) or \(TT\) occurs. Thus: \[P(\text{Rossa Wins}) = P(HH) + P(TT) = (2/3)^2 + (1/3)^2 = 5/9\]
Calculate the probability that the game will last for a full three rounds.
We need to calculate the probability of a tie after two rounds. There are two ways that a tie could occur: either Rossa wins the first round while Rodrigo wins the second, or Rodrigo wins the first round while Rossa wins the second. These two events are mutually exclusive and the probability of each is \(5/9 \times 4/9 = 20/81\) since successive coin flips are independent. Thus, the desired probability is \(40/81\).
Calculate the probability that Rodrigo will win the game.
Rodrigo needs to win two rounds to win the game. There are three ways this can happen. First, Rodrigo could win both rounds 1 and 2, in which case no third round is played. The probability of this event is \(4/9 \times 4/9 = 16/81\). Second Rodrigo could lose round 1 but win rounds 2 and 3. The probability of this event is \(5/9 \times 4/9 \times 4/9 = 80/729\). Finally, Rodrigo could lose round 2 but win rounds 1 and 3. The probability of this event is \(4/9 \times 5/9 \times 4/9 = 80/729\). Summing these probabilities, since their corresponding events are mutually exclusive, the probability that Rodrigo wins the game is \(304/729\approx 0.417\).
Yiwen is walking down the hallway and sees Rodrigo doing his victory dance: clearly Rossa has been defeated in spite of rigging the game. Given that Rodrigo won, calculate the probability that the game lasted for three rounds.
By the definition of conditional probability, \[\begin{equation*}
P(\text{3 Rounds}|\text{Rodrigo Won}) = \frac{P(\text{3 Rounds} \cap \text{Rodrigo Won})}{P(\text{Rodrigo Won})}
\end{equation*}\] We already calculated the denominator in the preceding part: it equals \(304/729\). To calculate the numerator we simply add up the probabilities of the two mutually exclusive ways in which Rodrigo could win in three rounds: (Win, Lose, Win) and (Lose, Win, Win). We calculated these probabilities in the preceding part: both were \(80/729\) so the numerator is \(160/729\). Taking the ratio of these gives \(160/304 \approx 0.526\). Given that Rodrigo won, it is slightly more likely than not that the game lasted for a full three rounds.
Bayes’ rule
Problem 18 Standard
The Triangle is a neighborhood that once housed a chemical plant but has become a residential area. Two percent of the children in the city live in the Triangle, and fourteen percent of these children test positive for excessive presence of toxic metals in the tissue. For children in the city who do not live in the Triangle, the rate of positive tests is only one percent. Let \(T\) be the event that a child lives in the Triangle and \(M\) be the event that a child tests positive for excessive presence of toxic metals.
If we randomly select a child who lives in the city, what is the probability that she both lives in the Triangle and tests positive?
By the multiplication rule: \[P(T\cap M) = P(M|T)P(T) = 0.14 \times 0.02 = 0.0028\]
If we randomly select a child who lives in the city, what is the probability that she tests positive?
By the law of total probability \[\begin{aligned}
P(M) &= P(M|T)P(T) + P(M|T^c)P(T^c)\\
&= 0.14 \times 0.02 + 0.01 \times 0.98\\
&= 0.0028 + 0.0098 \\
&= 0.0126
\end{aligned}\]
If we randomly select a child who lives in the city and she tests positive, what is the probability that she lives in the Triangle?
By Bayes’ Rule: \[\begin{aligned}
P(T|M) &= \frac{P(M|T)P(T)}{P(M)}\\
&= \frac{0.0028}{0.0126} = 2/9 \approx 0.22
\end{aligned}\]
Problem 19 Standard
From a past Oxford demonstration lecture.
Three percent of Tropicana brand oranges are already rotten when they arrive at the supermarket. In contrast, six percent of Sunkist brand oranges arrive rotten. A local supermarket buys forty percent of its oranges from Tropicana and the rest from Sunkist. Suppose we randomly choose an orange from the supermarket and see that it is rotten. What is the probability that it is a Tropicana?
Let \(T\) be the event that the orange is a Tropicana and \(R\) be the event that it is rotten. We are told that \(P(T) = 0.4\), so \(P(T^c) = 0.6\), along with \(P(R|T) = 0.03\) and \(P(R|T^c) = 0.06\). By the Law of Total Probability, \[P(R) = P(R|T)P(T) + P(R|T^c)P(T^c) = 0.03 \times 0.4 + 0.06 \times 0.6 = 0.012 + 0.036 = 0.048\] Hence, by Bayes’ Rule, \[P(T|R) = \frac{P(R|T)P(T)}{P(R)} = \frac{0.012}{0.048} = \frac{1}{4}\]
Problem 20 Standard
A startup is developing apps using three different operating systems: Microsoft Windows, Mac OSX, and Linux. On the first trial, apps compiled under Linux crash 10% of the time, compared to 20% of the time for Mac OSX and 30% of the time for Windows. Of the ten computers at the startup six run Linux, three run Mac OSX and one runs Windows. Sarah works at the startup and was randomly assigned one of these computers. Her app crashed on the first trial. Given this information, what is the probability that she was assigned a Windows machine? (Let \(C\) be the event that Sarah’s app crashes, \(W\) that she was assigned Windows, \(M\) Mac OSX and \(L\) Linux.)
We need to calculate \(P(W|C)\). By Bayes’ Rule \[P(W|C) = \frac{P(C|W)P(W)}{P(C)}\] By the law of total probability, \[\begin{aligned}
P(C) &= P(C|W)P(W) + P(C|M)P(M) + P(C|L)P(L)\\
&=0.3 \times 0.1 + 0.2 \times 0.3 + 0.1 \times 0.6\\
&= 0.03 + 0.06 + 0.06\\
&= 0.15
\end{aligned}\] Hence, \[P(W|C) = \frac{P(C|W)P(W)}{P(C)} = \frac{0.3 \times 0.1}{0.15}= 1/5 = 0.2\]
Problem 21 Standard
Someone in San Jose California, a city with 1 million residents, has stolen a car. The incident was captured on video, providing a detailed description of the thief. One out of every 10,000 people in San Jose fits this description. The police spot Amy walking down the street; she meets every detail of the description, so they arrest her. The only evidence against Amy is that she fits the description. The prosecutor argues as follows:
It is highly unlikely (far beyond a reasonable doubt) that an innocent person would fit this description. Hence, it is highly unlikely that Amy is innocent.
In parts (b)–(d) below, let \(I\) denote the event that someone from San Jose is innocent of the theft, and \(D\) be the event that she fits the description from the video.
How many of the residents of San Jose fit the description from the video?
There are 1 million people in San Jose. One out of every 10,000 fits the description. Thus, \(1000000/10000 = 100\) people fit the description.
Calculate \(P(I \cap D)\).
Of the 100 people in San Jose who fit the description, 99 are innocent and one is guilty. Hence, \(P(I\cap D) = 99/1000000\).
Calculate \(P(D|I)\).
There are 1 million people in San Jose, and only one of them is guilty. Hence \(P(I) = 999999/1000000\). By the definition of conditional probability, \[\begin{aligned}
P(D|I) = \frac{P(I \cap D)}{P(I)} = \frac{99/1000000}{999999/1000000} = \frac{99}{999999} \approx 1/10000
\end{aligned}\]
Calculate \(P(I|D)\).
From above, 100 out of 1 million people in San Jose fit the description. Thus, \(P(D) = 100/1000000\). By the definition of conditional probability, \[\begin{aligned}
P(I|D) = \frac{P(I \cap D)}{P(D)} = \frac{99/1000000}{100/1000000} = 99/100
\end{aligned}\]
In light of your responses to parts (c) and (d), evaluate the prosecutor’s argument. Should the jury vote to convict Amy? Explain briefly.
The jury should not vote to convict Amy. The prosecutor is correct that, given someone is innocent, it is highly unlikely that she will fit the description: this probability is \(P(D|I) \approx 1/10000\). The relevant probability, however, is the reverse conditional probability, namely the probability that someone is innocent given that she fits the description. This is \(P(I|D) = 99/100\) so it is highly likely that Amy is innocent. The prosecutor’s reasoning is fallacious. (FYI: this fallacy is commonly called the “prosecutor’s fallacy.”)
Problem 22 Standard
I have two six-sided dice in my pocket: one fair die and one loaded die. The fair die has the usual probabilities, but the probability of getting a 6 when rolling the loaded die is 1/2. Suppose I reach into my pocket and draw one of the two dice at random (both are equally likely to be drawn). I roll this randomly chosen die and get a 6. What is the probability that I drew the loaded die?
Let \(L\) be the event that I draw the loaded die, \(F\) be the event that I draw the fair die and \(6\) be the event that I roll a six. By Bayes’ rule, we have \[P(L|6) = \frac{P(6|L)P(L)}{P(6)}\] Calculating the denominator by the Law of Total Probability, we have \[\begin{aligned}
P(6) &= P(6|L)P(L) + P(6|F)P(F) \\
&= 1/2 \times 1/2 + 1/6\times 1/2\\
&= 1/4 + 1/12 \\
&= 1/3
\end{aligned}\] Hence, \[P(L|6) = \frac{1/4}{1/3} = 3/4 = 0.75\]
Problem 23 Standard
Sherlock Holmes has gone away on vacation, instructing Dr. Watson to water the flowers in his absence. Unfortunately Watson has a rather poor memory: the probability that he will remember to water the flowers is only 2/3. The flowers weren’t in the best shape when Holmes left: even if watered the probability that they will wither and die before Holmes returns is 1/2. If they aren’t watered, the probability that they will wither and die increases to 3/4. Holmes returns to find that his flowers have died. What is the probability that Watson forgot to water them?
By Bayes’ Rule: \[\begin{equation*}
P(\text{Forget}|\text{Die}) = \frac{P(\text{Die}|\text{Forget})P(\text{Forget})}{P(\text{Die})}
\end{equation*}\] We calculate the denominator using the law of total probability as follows: \[\begin{aligned}
P(\text{Die}) &= P(\text{Die}|\text{Forget})P(\text{Forget})+ P(\text{Die}|\text{Remember})P(\text{Remember})\\
&= 3/4 \times 1/3 + 1/2 \times 2/3 = 3/12 + 2/6 = 7/12
\end{aligned}\] Thus \(P(\text{Forget}|\text{Die}) = (3/12)/(7/12) = 3/7\). It is more likely than not that Watson remembered to water the flowers.
Problem 24 Standard
New England Patriots quarterback Tom Brady has a lucky coin, which he flips five times before every game. The coin is fair, so it has a probability 1/2 of coming up heads on any given flip.
What is the probability that Tom gets the same outcome for all 5 flips of his lucky coin? Express your answer as a fraction.
There are two possibilities here: Tom can either get 5 heads, or he can get 5 tails. Both have the same probability, \(\left(\frac{1}{2}\right)^5\). As they are mutually exclusive, we can add the probability of each possibility to get: \[P(5H \cup 5T) = P(5H) + P(5T) = \left(\frac{1}{2}\right)^5 + \left(\frac{1}{2}\right)^5 = \frac{1}{32} + \frac{1}{32} = \frac{1}{16}\]
Tom is a superstitious person. If (and only if) he gets 5 heads or 5 tails he will cheat by deflating the balls used in the game. If Tom cheats, the Patriots win with probability \(\frac{3}{4}\). But if Tom does not cheat, they win with probability \(\frac{2}{5}\). Given that the Patriots won the game, what is the probability that Tom cheated?
First write down all the probabilities given in the question statement. Let \(C\) be the event that Tom Brady cheats and \(W\) be the event that the Patriots win. (Using this notation \(C^c\) is the event that Tom doesn’t cheat.) From part (a), we know that \(P(C) = 1/16\) and therefore \(P(C^c) = 1 - P(C) = 15/16\). We are told in the problem statement that \(P(W|C) = 3/4\) while \(P(W|C^c) = 2/5\) and are asked to find \(P(C|W)\). By Bayes’ Rule: \[\begin{equation*}
P(\text{C}|\text{W}) = \frac{P(\text{C})P(\text{W}|\text{C})}{P(\text{W})}
\end{equation*}\] To calculate the denominator, we use the law of total probability: \[\begin{aligned}
P(\text{W}) &= P(\text{W}|\text{C})P(\text{C})+ P(\text{W}|\text{$C^c$})P(\text{$C^c$})\\
&= 3/4 \times 1/16 + 2/5 \times 15/16 \\
&= 3/64 + 3/8 = 3/64 + 24/64 = 27/64
\end{aligned}\] Now putting all of these pieces together, we get: \[\begin{equation*}
P(\text{C}|\text{W}) = \frac{P(\text{C})P(\text{W}|\text{C})}{P(\text{W})} = \frac{3/64}{27/64} = \frac{3}{27} = \frac{1}{9}
\end{equation*}\]
Problem 25 Standard
I have three boxes. The first contains two red balls, the second contains two blue balls, and the third contains one red ball and one blue ball. I choose a box at random so that each box is equally likely to be selected. I then reach inside and pull out a ball: it’s red. Calculate the probability that the other ball in the box is also red. Let \(B_1\) be the event that I chose the first box, \(B_2\) the second box, and \(B_3\) the third box. Let \(R\) be the event that I drew a red ball.
The problem statement specifies \(P(B_1) = P(B_2) = P(B_3) = 1/3\) along with \(P(R|B_1) = 1\), \(P(R|B_2) = 0\), and \(P(R|B_3) = 1/2\). We are asked to calculate the probability that the other ball is red given that the ball I drew is red. This is equivalent to asking for the probability of \(B_1\) given \(R\). By Bayes’ Rule \[P(B_1|R) = \frac{P(R|B_1)P(B_1)}{P(R)}\] For the denominator, \[\begin{aligned}
P(R) &= P(R|B_1)P(B_1) + P(R|B_2) P(B_2) + P(R|B_3) P(B_3) \\
&= 1 \times 1/3 + 0 \times 1/3 + 1/2 \times 1/3\\
&= 1/2
\end{aligned}\] Therefore, \(P(B_1|R) = (1/3) / (1/2) = 2/3\).
Problem 26 Standard
Bob is a randomly chosen resident of Peoria, a city in which 3% of people use cocaine. Bob tests positive for cocaine in a drug test that correctly identifies users 95% of the time and correctly identifies non-users 90% of the time. Calculate the probability that Bob is a cocaine user. (In your calculations, let \(U\) be the event that Bob is a cocaine user and \(T\) be the event that he tests positive.)
By the law of total probability: \[\begin{aligned}
P(T) &= P(T|U)P(U) + P(T|U^c)P(U^c) = \frac{95}{100} \times \frac{3}{100} + \frac{10}{100}\times \frac{97}{100}\\
&= (285 + 970) / (100 \times 100) = 1255 / (100 \times 100)
\end{aligned}\] Therefore, by Bayes’ rule \[P(U|T) = \frac{P(T|U)P(U)}{P(T)} = \frac{285}{1255} \approx 0.23\]
Problem 27 Standard
Of women who undergo regular mammograms, two percent have breast cancer. If a woman has breast cancer, there is a 90% chance that her mammogram will come back positive. If she does not have breast cancer there is a 10% chance that her mammogram will come back positive. Given that a woman’s mammogram has come back positive, what is the probability that she has breast cancer?
Let \(B\) be the event that a given woman has breast cancer and \(M\) be the event that her mammogram comes back positive. By Bayes’ Rule, \[P(B|M) = \frac{P(M|B)P(B)}{P(M)}\] By the law of total probability, \[\begin{aligned}
P(M) &= P(M|B)P(B) + P(M|B^c)P(B^c)\\
&= 0.9 \times 0.02 + 0.1 \times 0.98 = 0.018 + 0.098 = 0.116
\end{aligned}\] Hence, \[P(B|M) = \frac{0.9 \times 0.02}{0.116} = \frac{0.018}{0.116} \approx 0.16\]
Problem 28 Harder
From a past Oxford demonstration lecture.
Imagine there are two universities, University A and University B, who take different approaches to generating research findings. In University A, a team of well-informed experts develop a theory. Their theories are correct 90% of the time. Before publishing a theory, the scientists at University A do an experimental test of the theory to check whether it is correct (e.g. a clinical trial). The test is designed so that 90% of correct theories will pass the test and only 10% of false theories pass the test.
Let \(T\) be the event that the theory is true and \(\text{Pub}\) be the event that the theory passes the test and is published. Draw a Venn diagram to represent these probabilities.
Draw the sample space \(\Omega\) as a square with area 1, and let \(T\) cover 90% of it. Draw \(\text{Pub}\) as a second region overlapping \(T\). Together, \(T\) and \(\text{Pub}\) split the square into four regions. By the multiplication rule \(P(A \cap B) = P(B|A)P(A)\), their areas are \(P(T \cap \text{Pub}) = 0.9 \times 0.9 = 0.81\), \(P(T \cap \text{Pub}^c) = 0.9 \times 0.1 = 0.09\), \(P(T^c \cap \text{Pub}) = 0.1 \times 0.1 = 0.01\), and \(P(T^c \cap \text{Pub}^c) = 0.1 \times 0.9 = 0.09\). Almost all of \(\text{Pub}\) lies inside \(T\) (0.81 of its 0.82).
Calculate the probability that a theory from University A is published, i.e. \(P(\text{Pub})\).
By the Law of Total Probability, \[P(\text{Pub}) = P(\text{Pub}|T)P(T) + P(\text{Pub}|T^c)P(T^c) = 0.9 \times 0.9 + 0.1 \times 0.1 = 0.82\]
Calculate the probability that a published theory from University A is correct, i.e. \(P(T|\text{Pub})\).
By Bayes’ Rule, \[P(T|\text{Pub}) = \frac{P(\text{Pub}|T)P(T)}{P(\text{Pub})} = \frac{0.81}{0.82} \approx 0.99\]
In University B, a team of creative experts think of theories that would be rather surprising and interesting if true. These theories are correct only 5% of the time. Again, before publishing their theory, the scientists at University B do the same experimental test as in University A of the theory, i.e. the test is designed so that 90% of correct theories will pass the test and only 10% of false theories pass the test. Calculate the probability that a published theory from University B is correct.
Now \(P(T) = 0.05\). By the Law of Total Probability, \[P(\text{Pub}) = 0.9 \times 0.05 + 0.1 \times 0.95 = 0.045 + 0.095 = 0.14\] so by Bayes’ Rule, \[P(T|\text{Pub}) = \frac{0.045}{0.14} = \frac{9}{28} \approx 0.32\] The test is the same at both universities. Because University B’s theories are much less likely to be correct before testing, only about a third of its published theories are correct.
The Research Council governing the publication of research requires that published theories (i.e. those that pass the test) must be replicated before they can be used in practice. Assume that the replication test is like the first test, i.e. the test is designed so that 90% of correct theories will pass the test and only 10% of false theories pass the test. Compute the replicability rate of University A and University B (i.e. the pass rate of the second test).
Let \(\text{Rep}\) be the event that the theory passes the replication test. The problem tells us that \(P(\text{Rep}|T) = 0.9\) and \(P(\text{Rep}|T^c) = 0.1\). We also assume that, once we know whether a theory is true, the result of the replication test is independent of the result of the first test. Under this assumption, conditioning on \(\text{Pub}\) as well changes nothing, so \(P(\text{Rep}|T, \text{Pub}) = 0.9\) and \(P(\text{Rep}|T^c, \text{Pub}) = 0.1\). (Here \(P(\text{Rep}|T, \text{Pub})\) is shorthand for \(P(\text{Rep}|T \cap \text{Pub})\).) Probabilities conditional on \(\text{Pub}\) obey the same rules as ordinary probabilities, so the Law of Total Probability gives \[P(\text{Rep}|\text{Pub}) = P(\text{Rep}|T, \text{Pub})P(T|\text{Pub}) + P(\text{Rep}|T^c, \text{Pub})P(T^c|\text{Pub}) = 0.9 \times P(T|\text{Pub}) + 0.1 \times P(T^c|\text{Pub})\] For University A, part (c) gives \(P(T|\text{Pub}) = 81/82\), so \(P(T^c|\text{Pub}) = 1/82\) and the replication rate is \(0.9 \times \frac{81}{82} + 0.1 \times \frac{1}{82} = \frac{73}{82} \approx 0.89\). For University B, part (d) gives \(P(T|\text{Pub}) = 9/28\), so \(P(T^c|\text{Pub}) = 19/28\) and the replication rate is \(0.9 \times \frac{9}{28} + 0.1 \times \frac{19}{28} = \frac{5}{14} \approx 0.36\). Most of University A’s published theories pass the replication test. Almost two thirds of University B’s fail it, even though both universities use the same test.
Problem 29 Harder
Mars inc. produces the “colorful button-shaped chocolates” M&M’s. The contents of a bag of M&M’s changed in 1995 when tan M&M’s were replaced by blue M&M’s. The relative frequencies of the remaining colors changed as well. The values you will need to solve this problem appear in bold in the following table:
| Before 1995 (Old) |
– |
10% |
10% |
10% |
20% |
20% |
30% |
| After 1995 (New) |
24% |
– |
20% |
16% |
14% |
13% |
13% |
I have two bags of M&M’s: one from before 1995 (Old) and one from after 1995 (New). I randomly choose a bag, such that each is equally likely to be selected. I then make three independent random draws with replacement from the bag. I obtain: green, yellow, red. Given this information, what is the probability that I selected the Old bag? Let \(Old\) be the event that I drew the old bag and \(New\) be the event that I drew the new bag. Similarly, let \(G\) be the event that my first draw is green, \(Y\) that my second is yellow, and \(R\) that my third is red.
Here \(P(G,Y,R)\) is shorthand for \(P(G \cap Y \cap R)\). First, by the law of total probability, we have \[P(G,Y,R) = P(G,Y,R|Old)P(Old) + P(G,Y,R|New)P(New)\] Since I make my draws independently with replacement, \[\begin{aligned}
P(G,Y,R|Old) &= P(G|Old)P(Y|Old)P(R|Old)\\
&= 1/10 \times 1/5 \times 1/5 = 1/250
\end{aligned}\] and similarly \[\begin{aligned}
P(G,Y,R|New) &= P(G|New)P(Y|New)P(R|New)\\
&= 1/5 \times 14/100 \times 13/100 = 182/50000
\end{aligned}\] Thus, \[P(G,Y,R) = 1/250 \times 1/2 + 182/50000 \times 1/2\] and by Bayes’ Rule, \[\begin{aligned}
P(Old|G,Y,R) &= \frac{P(G,Y,R|Old)P(Old)}{P(G,Y,R)}\\
&= \frac{1/250 \times 1/2}{1/250 \times 1/2 + 182/50000 \times 1/2}\\
&= \frac{1/250}{1/250 + 182/50000} = \frac{200}{200 + 182}= 100/191 \approx 0.52
\end{aligned}\]
Problem 30 Harder
This question concerns a game played by rolling a fair, six-sided die with sides numbered 1–6. To play the game you roll the die once. Let \(x\) denote the number on the side that shows face-up. If \(x\) is even you win \(x\) dollars but if \(x\) is odd you win \(2x\) dollars. Now we change the rules of the game as follows: after rolling the die and observing \(x\) you are given the option to roll a second time. If you choose not to roll a second time, your winnings are calculated as before: \(x\) dollars if \(x\) is even and \(2x\) dollars if \(x\) is odd. If you do choose to roll again your winnings are calculated in the same way but based on whatever number comes up on your second roll. For example if your initial roll is 3 and you re-roll and get a 1 then you win 2 dollars.
Nina played this game once following the strategy of rolling a second time if and only if her first roll was a 1, 2, or 4. She won 10 dollars. Given this information, what is the probability that she chose to roll the die a second time?
Let \(A\) be the event that Nina chose to roll the die a second time and \(B\) be the event that she won 10 dollars. We are asked to calculate \(P(A|B)\). By Bayes’ Rule \[P(A|B)=\frac{P(B|A)P(A)}{P(B)}\] and by the Law of Total Probability \[\begin{aligned}
P(B) &= P(B|A)P(A) + P(B|A^c)P(A^c)
\end{aligned}\] Using this strategy, Nina will re-roll if she gets a 1, 2, or 4 on her initial roll. Thus, since the die is fair, \(P(A) = 1/2\) so we have \[P(A|B)=\frac{P(B|A)}{P(B|A) + P(B|A^c)}\] Now, \(P(B|A^c)\) is the probability that Nina won 10 dollars given that she did not roll twice. Since she is following this strategy, if Nina does not roll twice then she must have rolled a 3, 5, or 6 initially. Thus \(P(B|A^c)=1/3\). Now, recall that if Nina rolls twice then her winnings depend on the second roll only. This means that \(P(B|A)\) equals the probability of winning 10 dollars in the original version of the game, namely 1/6. Therefore \(P(A|B) = (1/6)/(1/3 + 1/6) = 1/3\). If Nina wins 10 dollars it is more likely than not that she only rolled once.
Problem 31 Harder
On my desk I have 10 cups: \(N_B\) of them are Blue Cups and the remaining \(10 - N_B\) are Red Cups. Each cup contains five balls: Blue Cups contain 4 blue balls and 1 red ball while Red Cups contain 4 red balls and 1 blue ball. I chose a cup at random so that each cup was equally likely to be selected. I then drew three balls at random with replacement from the chosen cup. In order, the balls I drew were: red, red, blue. Let \(C_B\) be the event that I chose a Blue Cup, \(C_R\) the event that I chose a Red Cup, and let \(RRB\) be the event that represents my three draws: a red ball, followed by another red ball, followed by a blue ball.
Suppose \(N_B\) is 5. Calculate \(P(C_B|RRB)\).
By the Law of Total Probability, \[\begin{aligned}
P(RRB) &= P(RRB|C_B)P(C_B) + P(RRB|C_R)P(C_R)\\
&= (1/5 \times 1/5 \times 4/5) \times 1/2 + (4/5 \times 4/5 \times 1/5) \times 1/2\\
&= 2/125 + 8/125 = 10/125
\end{aligned}\] Hence, by Bayes’ Rule, \[\begin{aligned}
P(C_B|RRB) &= \frac{P(RRB|C_B)P(C_B)}{P(RRB)}= \frac{2/125}{10/125} = 1/5
\end{aligned}\]
Now suppose that we do not know the value of \(N_B\). How large would \(N_B\) have to be for it to be more likely that I drew from a blue cup given that the event \(RRB\) has occurred? Prove your answer.
This is identical to the previous part with one change: now we have \(P(C_B) = N_B/10\) and \(P(C_R) = 1 - N_B/10\) rather than 1/2. Hence, the calculation for the Law of Total Probability becomes \[\begin{aligned}
P(RRB) &= (4/125) \times (N_B/10)+ (16/125) \times(10 - N_B)/10\\
&= \left[4N_B + 16(10 - N_B) \right]/(1250)\\
&= (160 - 12N_B)/1250
\end{aligned}\] and similarly for Bayes’ Rule \[\begin{aligned}
P(C_B|RRB) &= \frac{4N_B/1250}{(160 - 12N_B)/1250} = \frac{4N_B}{160 - 12N_B} = \frac{N_B}{40 - 3N_B}
\end{aligned}\] We need the smallest \(N_B\) such that this quantity is greater than 1/2: \[\begin{aligned}
N_B/(40 - 3N_B) &> 1/2\\
2 N_B &> 40 - 3N_B\\
N_B &> 8
\end{aligned}\] Therefore \(N_B\) would have to be at least 9.
Suppose that I made my draws without replacement. What is \(P(C_B|RRB)\) in this case? Briefly explain your answer.
If we draw without replacement, then getting two red balls makes it impossible that we’re drawing from a Blue Cup, since Blue Cups only have one red ball. Hence \(P(C_B|RRB) = 0\).
Problem 32 Challenge
From the HT 2023 collection.
In the land of Wendimoor, one in fifty people have a disease called pararibulitis that causes painful and disturbing hallucinations. Fortunately this condition is easily treated by a daily dose of Subtracterall. Because Subtracterall has some mildly unpleasant side effects, the physicians of Wendimoor only prescribe it for patients who test positive for pararibulitis. The test has an accuracy of 80%. In other words, a person with pararibulitis has an 80% chance of testing positive and a person without pararibulitis has an 80% chance of testing negative.
Amanda is a randomly-chosen citizen of Wendimoor. We test her for pararibulitis and the test comes back positive. What is the probability that she has the disease? In your solution, let \(T\) be the event that Amanda tests positive and \(D\) be the event that she has the disease.
By Bayes’ Rule: \[\mathbb{P}(D|T) = \frac{\mathbb{P}(T|D)\mathbb{P}(D)}{\mathbb{P}(T)}\] And by the law of total probability, \[\begin{aligned}
\mathbb{P}(T) &= \mathbb{P}(T|D)\mathbb{P}(D) + \mathbb{P}(T|D^c) \mathbb{P}(D^c) \\
&= 4/5 \times 1/50 + 1/5 \times 49/50\\
&= 4/250 + 49/250\\
&= 53/250
\end{aligned}\] Therefore: \(\mathbb{P}(D|T) = (4/250) / (53/250) = 4/53 \approx 0.075\). Just for fun, here’s a much faster way of solving the problem by calculating the odds: \[\frac{\mathbb{P}(D|T)}{\mathbb{P}(D^c|T)} = \frac{\mathbb{P}(T|D)}{\mathbb{P}(T|D^c)}\cdot\frac{\mathbb{P}(D)}{\mathbb{P}(D^c)} = \frac{4/5}{1/5} \cdot \frac{1/50}{49/50} = \frac{4}{49}.\] Expressed as a probability, this becomes \(4/(49 + 4) = 4/53\) as above.
Half of people with pararibulitis notice that they have symptoms and visit a physician to get tested. At the same time, one in ten of people who do not have pararibulitis mistakenly believe that they have symptoms and visit a physician to get tested. Anyone who visits a physician and tests positive gets a prescription; no one else does. Among people with pararibulitis, what fraction have a prescription? Among people who do not have pararibulitis, what fraction have a prescription? Let \(S\) be the event that someone is prescribed Subtracterall and \(V\) be the event that she visits a physician to get tested.
By the law of total probability, \[\mathbb{P}(S|D) = \mathbb{P}(S|D,V)\mathbb{P}(V|D) + \mathbb{P}(S|D,V^c)\mathbb{P}(V^c|D).\] Here \(\mathbb{P}(S|D,V)\) is shorthand for \(\mathbb{P}(S|D \cap V)\). To simplify this expression, first notice that \(\mathbb{P}(S|D,V^c)=0\). In words: if you don’t visit a physician to get tested, you can’t get a prescription even if you have the disease. Next, notice that \(\mathbb{P}(S|D,V)=\mathbb{P}(T|D)=4/5\). In words: if you visit a physician to get tested and you have the disease, the chance that you will get a prescription equals the chance that you test positive. Therefore: \[\mathbb{P}(S|D) = \mathbb{P}(T|D)\mathbb{P}(V|D) = 4/5 \times 1/2 = 2/5.\] Similarly, \[\begin{aligned}
\mathbb{P}(S|D^c) &= \mathbb{P}(S|D^c,V)\mathbb{P}(V|D^c) + \mathbb{P}(S|D^c,V^c) \mathbb{P}(V^c|D^c)\\
&= \mathbb{P}(T|D^c) \mathbb{P}(V|D^c) = 1/5 \times 1/10 = 1/50.
\end{aligned}\]
Continuing from part (b), Amanda argues that pararibulitis is under-diagnosed: most people who have the disease are not prescribed the Subtracterall. Todd argues that pararibulitis is over-diagnosed: most people with a prescription for Subtracterall do not in fact have pararibulitis. Who’s right? Explain briefly.
They’re both right! Surprisingly, it’s possible for a disease to be both over-diagnosed and under-diagnosed. From our solution to the previous part, we know that only 40% of people with pararibulitis are prescribed Subtracterall; most of the people with this disease are untreated. At the same time, Todd is right that most people who take Subtracterall do not have pararibulitis. By Bayes’ Rule \[\mathbb{P}(D|S) = \frac{\mathbb{P}(S|D)\mathbb{P}(D)}{\mathbb{P}(S)} = \frac{2/5 \times 1/50}{\mathbb{P}(S)}\] and by the law of total probability, \[\begin{aligned}
\mathbb{P}(S) &= \mathbb{P}(S|D)\mathbb{P}(D) + \mathbb{P}(S|D^c)\mathbb{P}(D^c)\\
&= 2/5 \times 1/50 + 1/50 \times 49/50\\
&= 1/50 \times 69/50
\end{aligned}\] Therefore, \(\mathbb{P}(D|S) = (2/5) / (69/50) = 20/69 \approx 0.29\).
As above, it’s simpler to solve this using odds: \[\frac{\mathbb{P}(D|S)}{\mathbb{P}(D^c|S)} = \frac{\mathbb{P}(S|D)}{\mathbb{P}(S|D^c)} \cdot \frac{\mathbb{P}(D)}{\mathbb{P}(D^c)} = \frac{2/5}{1/50} \cdot \frac{1/50}{49/50} = \frac{20}{49}\] from which we immediately see that it’s much less likely than not that a person who takes Subtracterall has pararibulitis. (We don’t even need to convert to a probability to see this because 20 is much smaller than 49.)
Problem 33 Challenge
From the HT 2024 collection.
Stephen has two urns, each of which contains three balls. The “blue urn” contains two blue balls and one red ball; the “red urn” contains two red balls and one blue ball. Neither urn has any distinguishing characteristics. Without looking inside to see which balls they contain, there is no way to tell the urns apart. To determine whether Frank, Natalie and Ruth should be allowed to continue teaching undergraduates, Stephen decides to assess their understanding of conditional probability. He flips a fair coin to choose randomly between the two urns. Then he lets Frank, Natalie and Ruth make one draw from the urn apiece. They’re not allowed to peek inside the urn, and they have to replace their ball after making the draw. They’re also not allowed to show each other which ball they drew. After each tutor makes their draw, they state aloud which urn they think Stephen has chosen.
Frank goes first: he draws a blue ball so he guesses that Stephen has chosen the blue urn. Based on the information available to him, what is the probability that Frank’s guess is correct?
By Bayes’ Rule: \[\mathbb{P}(\text{Blue Urn}|\text{Blue Ball}) = \frac{\mathbb{P}(\text{Blue Ball}|\text{Blue Urn})\mathbb{P}(\text{Blue Urn})}{\mathbb{P}(\text{Blue Ball})}\] and by the Law of Total Probability: \[\begin{aligned}
\mathbb{P}(\text{Blue Ball}) &= \mathbb{P}(\text{Blue Ball}|\text{Blue Urn})\mathbb{P}(\text{Blue Urn}) + \mathbb{P}(\text{Blue Ball}|\text{Red Urn})\mathbb{P}(\text{Red Urn})\\
&= 2/3 \times 1/2 + 1/3 \times 1/2.
\end{aligned}\] Therefore, \[\mathbb{P}(\text{Blue Urn}|\text{Blue Ball}) = \frac{2/3 \times 1/2}{2/3 \times 1/2 + 1/3 \times 1/2} = \frac{2/3}{2/3 + 1/3} = 2/3.\]
Natalie goes second. She knows that Frank is rational: since he guessed “blue urn” this means he must have seen a blue ball. She reaches into the urn and draws a red ball. Show that, based on all the information available to Natalie, the probability that Stephen chose the blue urn is \(1/2\).
To keep the notation from getting out of control, let \(U_B\) be the event that Stephen chose the blue urn, \(U_R\) be the event that Stephen chose the red urn, \(F_B\) be the event that Frank draws a blue ball, and \(N_R\) be the event that Natalie draws a red ball. By Bayes’ Rule \[\mathbb{P}(U_B|F_B \cap N_R) = \frac{\mathbb{P}(F_B\cap N_R|U_B)\mathbb{P}(U_B)}{\mathbb{P}(F_B \cap N_R)}.\] Since Frank’s draw is independent of Natalie’s given the urn–they’re different people making different draws from the same urn at different points in time–the Law of Total Probability gives \[\begin{aligned}
\mathbb{P}(F_B\cap N_R) &= \mathbb{P}(F_B \cap N_R|U_B) \mathbb{P}(U_B) + \mathbb{P}(F_B \cap N_R|U_R)\mathbb{P}(U_R) \\
&= 2/3 \times 1/3 \times 1/2 + 1/3 \times 2/3 \times 1/2 \\
&= 1/9 + 1/9
\end{aligned}\] and therefore we obtain \[\mathbb{P}(U_B|F_B \cap N_R) = \frac{1/9}{1/9 + 1/9} = 1/2.\]
Natalie guesses “blue urn.” Ruth knows that both Frank and Natalie are rational. Since Frank guessed “blue urn”, he must have drawn a blue ball. But she can’t be sure which color ball Natalie drew. Natalie always guesses the urn that has the higher probability based on the information available to her. In the event of a tie, she guesses randomly. Since Natalie guessed “blue urn” Ruth realizes that there are two possibilities: either Natalie drew a blue ball, or she drew a red ball and randomly guessed “blue urn.” Ruth reaches into the urn and draws a red ball. What should Ruth guess? Find the probability that her guess is correct, based on the information available to her.
This one is tricky! Start by asking what Ruth knows for sure: she knows that Frank drew a blue ball and that she herself drew a red one. By exactly the same reasoning as in the preceding part, Ruth’s posterior probability of \(U_B\) is \(1/2\), based on her own draw and what she knows about Frank’s draw. Notice that this is exactly the same as the prior probability that Stephen chose the blue urn. Since Ruth’s draw and Frank’s draw precisely “cancel each other out” in terms of the information they provide, we can ignore them when solving the problem. So we pretend that Frank and Ruth didn’t draw any balls. Ruth knows that Natalie drew a ball and guessed “blue urn”. Ruth also knows that Natalie’s decision rule is to guess “blue urn” if she draws a blue ball, and to guess randomly if she draws a red ball. Abbreviate “Natalie guessed blue” by \(\text{NGB}\). By Bayes’ Rule: \[\mathbb{P}(U_B|\text{NGB}) = \frac{\mathbb{P}(\text{NGB}|U_B)\mathbb{P}(U_B)}{\mathbb{P}(\text{NGB})}.\] By the Law of Total Probability, \[\mathbb{P}(\text{NGB}) = \mathbb{P}(\text{NGB}|U_B)\mathbb{P}(U_B) + \mathbb{P}(\text{NGB}|U_R)\mathbb{P}(U_R).\] We know that \(\mathbb{P}(U_B) = \mathbb{P}(U_R) = 1/2\), so Bayes’ Rule simplifies to: \[\mathbb{P}(U_B|\text{NGB}) = \frac{\mathbb{P}(\text{NGB}|U_B)}{\mathbb{P}(\text{NGB}|U_B) + \mathbb{P}(\text{NGB}|U_R)}.\] It remains to calculate \(\mathbb{P}(NGB|U_B)\) and \(\mathbb{P}(\text{NGB}|U_R)\). Each of these requires a second application of the Law of Total Probability, in which \(\mathbb{P}(\text{NGB}|U_B, N_B)\) is shorthand for \(\mathbb{P}(\text{NGB}|U_B \cap N_B)\): \[\begin{aligned}
\mathbb{P}(\text{NGB}|U_B) &= \mathbb{P}(\text{NGB}|U_B, N_B)\mathbb{P}(N_B|U_B) + \mathbb{P}(\text{NGB}|U_B,N_R)\mathbb{P}(N_R|U_B)\\
&= 1 \times \mathbb{P}(N_B|U_B) + 1/2 \times \mathbb{P}(N_R|U_B)\\
&= 2/3 + 1/2 \times 1/3 = 5/6 \\ \\
\mathbb{P}(\text{NGB}|U_R) &= \mathbb{P}(\text{NGB}|U_R, N_B)\mathbb{P}(N_B|U_R) + \mathbb{P}(\text{NGB}|U_R,N_R)\mathbb{P}(N_R|U_R)\\
&= 1 \times \mathbb{P}(N_B|U_R) + 1/2 \times \mathbb{P}(N_R|U_R)\\
&= 1/3 + 1/2 \times 2/3 = 4/6.
\end{aligned}\] Therefore, \(\mathbb{P}(U_B|\text{NGB}) = \displaystyle\frac{5/6}{4/6 + 5/6} = 5/9\). Since this is bigger than \(1/2\), Ruth should guess “blue urn”.
Now suppose that Frank, Natalie, and Ruth had some way to surreptitiously inform one another about which balls each of them had drawn, without Stephen realizing. As before, Frank draws a blue, Natalie red, and Ruth red. But now Ruth knows exactly which ball each of them drew. What should Ruth guess? Find the probability that her guess is correct based on the information available to her in this alternative scenario. Note that you can solve this part without solving part (c).
As above, Frank and Ruth’s draws “cancel out” so we can ignore them: we’re in exactly the same situation as at the start of the problem, where we have a prior probability of \(1/2\) for each urn. But now Ruth also knows for sure that Natalie drew a red ball. So by the same reasoning as in part (a), with red replacing blue, the posterior probability that Stephen chose the red urn is \(2/3\). Ruth should guess red.