Probability please help

euler42 · #1 $Old$ November 6th, 2009, 03:52 PM

I have two probability story problems that I can't figure out. Here they are:

Mrs. Miller manages a restaurant which is currently hiring. On Friday she interviewed 6 dishwashers, 1 line cook and 5 waiters. On Saturday she interviewed 3 dishwashers, 3 line cooks and 6 waiters for employment. Each day, one woman applied for a job, while the rest of the applicants were men. How probable is it that at least one of the women was a dishwasher?

On Monday this week Mr. Munton had a meeting at the Cultural Center, and on Friday he had another meeting. Of the conference rooms which his company uses, 3 face North, 3 face South, 2 face East and 4 face West. All the rooms receive equal use. How probable is it that although he went to meetings on both Monday and Friday, Mr. Munton only saw the view from one side of the Cultural Center?

Please show me which probability formulas you used and a detailed example of how you got the answer you did. Thanks.

Grandad · #2 $Old$ November 8th, 2009, 12:01 AM

Hello euler42

Welcome to Math Help Forum!

Quote:

Originally Posted by euler42 $View Post$

I have two probability story problems that I can't figure out. Here they are:

Mrs. Miller manages a restaurant which is currently hiring. On Friday she interviewed 6 dishwashers, 1 line cook and 5 waiters. On Saturday she interviewed 3 dishwashers, 3 line cooks and 6 waiters for employment. Each day, one woman applied for a job, while the rest of the applicants were men. How probable is it that at least one of the women was a dishwasher?

On Monday this week Mr. Munton had a meeting at the Cultural Center, and on Friday he had another meeting. Of the conference rooms which his company uses, 3 face North, 3 face South, 2 face East and 4 face West. All the rooms receive equal use. How probable is it that although he went to meetings on both Monday and Friday, Mr. Munton only saw the view from one side of the Cultural Center?

Please show me which probability formulas you used and a detailed example of how you got the answer you did. Thanks.

1) On Friday, 6 out of 12 interviewees were dishwashers. If one of these 12 is chosen at random as the one woman, the probability that she is a dishwasher is $\tfrac{6}{12} = \tfrac12$ .

Similarly the probability that the woman is a dishwasher on Saturday is $\tfrac{3}{12}=\tfrac14$ .

The simplest way to calculate the probability that at least one of these events occurred is to calculate the probability that neither of them did, and subtract the result from 1. So, the probability that the woman was not a dishwasher on Friday = $1 - \tfrac12=\tfrac12$ ; and the probability that the woman was not a dishwasher on Saturday was $1 - \tfrac14 = \tfrac34$ . Therefore the probability that neither of these happened is $\tfrac12\times \tfrac34 = \tfrac 38$ .

So the probability that at least one was a dishwasher is $1 - \tfrac38 = \tfrac58$ .

2) The probability that he faced North on both days is $\tfrac14\times\tfrac14 = \tfrac{1}{16}$ .

The probability that he faced East on both days is $\tfrac16\times\tfrac16 =\tfrac{1}{36}$ .

Similarly the probability that he faced South on both day is ...? ...and West ...?

These four events are mutually exclusive (i.e. if one of them occurs, then none of the others occurs as well). So we can add their probabilities together to find the probability that one of these occurred. I'll leave the rest to you.

Grandad