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Mary remembers to take her purse with her only 3/4 of the time when she brings it into a store. She visits three stores in succession. She realizes that she had it when she entered the first store, but doesn't have it when she leaves the third. Find the probability that the umbrella is in each of the stores.

So after leaving the first store, the probability that it is in store 1 is 1/4. After leaving the second store, I calculated the probability that she had the purse with her after leaving the first floor, multiplied with the probability that she had the purse as she leaves the second store, which is 3/4 x 1/4 = 3/16. Is this method correct?

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  • $\begingroup$ 1/4 for the first store. (3/4)^2 it is in the third store. And 1-1/4 -(3/4)^2 it is in the second store. $\endgroup$ – Winther May 9 '14 at 0:10
  • $\begingroup$ @Winther: that's not right at all, see below. That gives probabilities of $\frac{4}{16}, \frac{3}{16}, \frac{9}{16}$ for stores 1, 2, 3, respectively. $\endgroup$ – GMB May 9 '14 at 0:18
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The method is almost correct, but it needs some tweaking.

You'll notice that your method gives you $\frac{1}{4}$ for store 1, $\frac{3}{16}$ for store 2, and $\frac{9}{64}$ for store 3. But these don't sum up to $1$! The remaining probability corresponds to the probability that she actually kept her umbrella throughout - but you've been told that's not the case.

You need to condition on the fact that she leaves the umbrella in one of the stores. The fix is easy: just throw away the probability corresponding to her keeping the umbrella, and then rescale your remaining values to add up to $1$.

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  • $\begingroup$ Sorry, I'm still a little confused. Would the probability that she keeps the umbrella throughout be (3/4)^3? Because then 1/4, 3/16, 9/64, and (3/4)^3 add up to 1. $\endgroup$ – user148790 May 9 '14 at 0:14
  • $\begingroup$ Yes, that's exactly right -- does that clarify things? $\endgroup$ – GMB May 9 '14 at 0:16
  • $\begingroup$ yes - thank you very much! $\endgroup$ – user148790 May 9 '14 at 0:30

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