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The problem states that a typical customer buys the bread $60\%$ of the time and fruit $50\%$ of the time on each visit. Also the probability that the customers buy both bread and fruit is $0.3$. Then, what is the probability that a typical customer buys Bread Exactly Three times in the next Five visits?

To solve the problem, I set the given conditions like this:

$P(B)=0.6$, $P(F)=0.5$, and $P(B \,and\, F)=0.3$. Then from here, I found that $P(B\,or F\,)=0.8$.

But after this, I couldn't come up with the right solution. Can you help me please? THanks!

This problem is #7 in part 2 of [this test] http://www.wsmc.net/contests/2008_Contest/regtopprob.pdf

Free Spirit
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  • When you have a case where an event must occur some exact amount of times, using binomial probability as JPi did is often your best choice. – MT_ Feb 26 '14 at 04:19

1 Answers1

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Assuming that what the consumer buys at one visit is independent of what she buys at any other, you can use a binomial, i.e. the probability would be

$${5\choose 3} 0.8^30.2^2$$

JPi
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  • Yeah,,,actually that was what I did initially and I got $0.2038$ for that. But the solution says it is $0.345$ and that's why I'm stuck :( – Free Spirit Feb 26 '14 at 04:24
  • Looks like the solution works out the probability that a consumer only buys bread on three out of the five times. The numbers work exactly. – JPi Feb 26 '14 at 04:29
  • SO would it be $0.3^30.8^25C3$? But I need to multiply 2 to get the right answer. Where did I miss 2? – Free Spirit Feb 26 '14 at 05:43