Question

2. Find the following probabilities.
(a) Suppose 75% of women live past age 65 and 55% of women live past age
80. What is the probability that a woman who is now 65 will live past 80?
(b) What’s the probability that a roll of two fair six-sided dice will sum to 6?
What’s the probability of this once you observe that one of the dice came
up 2?
(c) A fair six-sided dice is tossed twice. The second toss is higher than the
first. What’s the probability that the first toss was at least 4?

4. Recall the cookie problem from lecture. We have two bowls, Bowl 1 and Bowl 2. Bowl 1 contains 25%
chocolate and 75% vanilla cookies; Bowl 2 has 50% of each. For this problem, assume each bowl is large
enough that drawing a single cookie does not appreciably alter this ratio.
Suppose we draw two cookies from the bowl and they are both chocolate. Calculate the posterior
probabilities of the two bowls in two ways:
(a) by treating the two cookies as one simultaneous piece of evidence
(b) by updating the prior probabilities once using the rst chocolate cookie, and using the posterior
probabilities as prior probabilities in a second update.

5. Suppose instead we draw two cookies; one is chocolate and the other is vanilla. Calculate the posterior
probabilities. Does it matter which cookie we drew rst? Why or why not?

Answer 1

2(a)

From the given information we have

P(live past age 65) = 0.75

P(live past age 65 and live past age 80) = 0.55

The probability that a woman who is now 65 will live past 80 is

P(live past age 80 | live past age 65) = P(live past age 65 and live past age 80) / P(live past age 65) = 0.55 / 0.75 = 0.7333

Answer: 0.7333

(b)

Following is the possible outcomes when we sum the outcomes of rolls of two dice:

Out of 36 outcomes, 5 shows sum 6 so the probability that a roll of two fair six-sided dice will sum to 6

P(sum 6) = 5 / 36 = 0.1389

Out of 5 outcomes showing sum 6, in two outcomes dice has 2 so

P(one die is 2| sum 6) = 2/5 = 0.40

(c)

Out of 15 outcomes with  second toss is higher than the first, 3 has first toss at least 4 so

P(first at least 4 | second is higher) = 3 /15 = 0.20