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Discrete Probability Worksheet 2: Conditional Probability
- Suppose there is a test for checking the presence of skin cancer. When cancer is present, the test is positive 90% of the time and negative the other 10%. When cancer is not present, the test is positive 10% of the time, and negative the other 90%. Furthermore, the probability of having cancer is 1%. If someone receives the test and the result is positive, what is the probability that they have cancer? Hint: Use Bayes’ theorem.
Solution: Let Cancer denote the event that the person has cancer and let Positive de- note the event that the person’s test result is positive. We can summarize the information given in the problem as
- P (Cancer) = 0. 01
- P (Positive | Cancer) = 0. 9
- P (Positive | CancerC^ ) = 0. 1
We are trying to find P (Cancer | Positive). Using Bayes’ theorem, we have
P (Cancer | Positive) = P (Positive | Cancer)P (Cancer) P (Positive)
P (Positive)
We can find the denominator above as follows:
P (Positive) = P (Positive ∩ Cancer) + P (Positive ∩ CancerC^ ) = P (Positive | Cancer)P (Cancer) + P (Positive | CancerC^ )P (CancerC^ ) = 0. 9 · 0 .01 + 0. 1 · 0. 99
So the final answer is (^0). 9 · 00 .01+0.^9 ·^0.^01. 1 · 0. 99 ≈ 0 .0833. In other words, even if someone tests positive, there is still less than a 10% chance they have cancer. This is because the initial probability of having cancer is so low.
- Suppose that there are two slot machines, one of which pays out 10% of the time and the other pays out 20% of the time. Unfortunately, you have no idea which is which. Suppose you randomly choose a machine and put in a quarter. If you don’t get a jackpot, what is the chance that you chose the machine that pays out 20% of the time? If you had instead gotten a jackpot, what would be the chance that you chose the one that pays out 20% of the time?
Solution: This is very similar to the previous problem. Let S be the event that you chose the first slot machine, i.e. the one that pays out 10% of the time (so SC^ is the event that you chose the second slot machine). Let J be the event that you got a jackpot. The information given in the problem is
• P (J | S) = 0. 1
• P (J | SC^ ) = 0. 2
We want to find P (SC^ | JC^ ) and P (SC^ | J). We can find both by applying Bayes’ theorem, just as in the previous problem (though this time we won’t explain every detail of the calculation):
P (SC^ | JC^ ) =
P (JC^ | SC^ )P (SC^ )
P (JC^ )
P (JC^ | SC^ )P (SC^ )
P (JC^ | SC^ )P (SC^ ) + P (JC^ | S)P (S)
Similarly,
P (SC^ | J) =
P (J | SC^ )P (SC^ )
P (J)
P (J | SC^ )P (SC^ )
P (J | SC^ )P (SC^ ) + P (J | S)P (S)
- Kidney stones is an affliction that comes in two varieties: small stones and large stones. Suppose that there are two treatments for kidney stones: treatment A and treatment B. Suppose that the success probabilities of these two types of treatment are as shown in the following table. Treatment A Treatment B Small Stones 93% 87% Large Stones 73% 68% Also suppose that a patient with kidney stones is equally likely to have small stones or large stones and that patients with small stones receive treatment A with probability 20% and patients with large stones receive treatment A with probability 80%. All patients who don’t receive treatment A receive treatment B. Given that a patient receives treatment A, what is the chance that it is successful? Given that a patient receives treatment B, what is the chance that it is successful? Which treatment do you think is better? By the way, this is a real example. The general phenomenon is known as “Simpson’s paradox.”
Comment: If you only see these final answers, it is tempting to believe that treatment B is better. However, when you look at the table of success probabilities above, it seems that treatment A is better for both small stones and large stones! What is going on? The answer is that despite that fact that treatment A is apparently more successful, treatment A is used more frequently for patients with large stones. But both treatments A and B are less likely to succeed on patients with large stones. Here’s one way to think about this: treatment A is most often used on patients for which both methods have trouble, whereas treatment B is most often used on patients for which both methods work well, creating the illusion that treatment B is better overall. The lesson here is not that treatment A is better—although based on the evidence given above, that seems likely—but rather that you should be careful when reasoning about probabilities: they don’t always indicate what you intuitively think they should.
- Show that your belief in something should never increase both when some other event occurs and when it doesn’t occur. Formally, show that if P (A | B) > P (A) then P (A | Bc) < P (A). By the way, this may seem like an obvious fact, but it has some surprising implications. For instance, if you believe Bitcoin might be a bubble and if the price of Bitcoin rising increases your belief that Bitcoin was a bubble then the price of Bitcoin falling should decrease your belief that Bitcoin was a bubble (and vice-versa).
Solution: Let’s assume for a moment that P (A | Bc) ≥ P (A) and see what that implies. As we’ve seen in class, we can always write
P (A) = P (A ∩ B) + P (A ∩ Bc).
By definition of conditional probability, we also have that
P (A ∩ B) = P (A | B)P (B) P (A ∩ Bc) = P (A | Bc)P (Bc).
Therefore P (A) = P (A | B)P (B) + P (A | Bc)P (Bc). But since we have assumed that P (A | B) > P (A) and P (A | Bc) ≥ P (A), this means that P (A) > P (A)P (B) + P (A)P (Bc) = P (A)(P (B) + P (Bc)). By the laws of probability, P (B) + P (Bc) = 1. Therefore the above equation implies that
P (A) > P (A).
Since this conlcusion is absurd, our original assumption must have been false. So we can conclude that P (A | Bc) < P (A).
- Suppose you are playing a game where someone rolls two fair 6-sided dice. If both rolls are ones, you win a million dollars. (a) If you are told that the first roll is a one, what is the chance that you will win?
Solution: Let A be the event that the first roll is a one and B the event that both rolls are one. The question is asking for P (B | A). By definition,
P (B | A) =
P (B ∩ A)
P (A)
The intersection of A and B is just B—if both rolls are one then the first roll must be one. The probability that both rolls are one is
P (B) =
and the probability that the first roll is one is
P (A) =
Therefore the answer is 1 / 62 1 / 6
(b) If you are told that at least one of the rolls is a one, what is the chance that you will win?
Solution: Let C be the event that at least one roll is a one and let B be as in the solution to part (a). The question is asking for P (B | C). Once again by definition of conditional probability, P (B | C) =
P (B ∩ C)
P (C)
As before, B ∩ C = B since if both rolls are one, then at least one roll is one. So we just need to calculate P (C). To do this, let’s define our sample space Ω to be the set of all pairs of numbers between one and six. Since the dice are fair, every outcome in Ω is equally likely. And |Ω| = 36. To find |C|, observe that the number of pairs of numbers that don’t contain a one is 5 × 5 = 25. So |C| = |Ω| − 25 = 11. Therefore
P (C) =
and so P (B | C) =
