Tried this problem and not getting the answer.

Out of 100 customers, 52 buy a shirt, 44 buy a tie and 30 buy both. Determine the probability that a customer who has bought a shirt will also buy a tie?

ANS = .58.

So the end it states SHIRT AND TIE?? so it's a multiplication probability?

I was using P(A and B) = P(A) + P(B) - P(A and B)

.52 + .44 - (.3) = .66

BUT MY guess that is wrong. I think it's a two step process. BUT have no clue.

Any help would be apprecaited. Thanks

Jo

Re: Tried this problem and not getting the answer.

Quote:

Originally Posted by

**bradycat** Determine the probability that a customer who has bought a shirt will also buy a tie?

Are you asking if you need to determine the probability, as in "Do I really need to determine... ?"? (Smile)

Quote:

Originally Posted by

**bradycat** So the end it states SHIRT AND TIE?? so it's a multiplication probability?

No, it's conditional probability.

Quote:

Originally Posted by

**bradycat** I was using P(A and B) = P(A) + P(B) - P(A and B)

The left-hand side should be P(A or B), but this equality is not relevant to this problem. Use the definition of conditional probability: P(A | B) = P(A and B) / P(B). In fact, this is the familiar formula for probability: the number of successful outcomes divided by the total number of outcomes. The difference is that in computing the probability of an event A provided a different event B is known to occur, the sample space (the set of outcomes) is restricted to outcomes in B. Here, instead of all customers, the new sample space consists of only those customers that bought a shirt, and the successful outcomes are those who in addition to a shirt also bought a tie.

Re: Tried this problem and not getting the answer.

Quote:

Originally Posted by

**bradycat** Out of 100 customers, 52 buy a shirt, 44 buy a tie and 30 buy both. Determine the probability that a customer who has bought a shirt will also buy a tie?

ANS = .58.

You are asked to find $\displaystyle P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{0.3}{0.52}$.

Re: Tried this problem and not getting the answer.

I was thinking that formula, but thought it was wrong. Thank you for the help. I understand it now.

The wording for me is tricky, as I have a technology math mind, and this stats sure throws me for a loop on different thinking.