# Probability of receiving bonus

1. Jul 25, 2014

### desmond iking

1. The problem statement, all variables and given/known data
A factory of 36 male workers and 64 female workers. woth 10 male workers earning less than $1000 per month and 17 female workers earning at least$1000 per month . At the end of year, workers earning less than $1000 are given a bonus of$1000 whereas others receive a month's salary.
If a male worker and a female worker are randomly chosen , find the probability that exactly one worker receive one month salary
here's my working.

P(male receive one month salary , female not receive one month salary) +P(female not receive one month salary , male receive one month salary) + P(female receive one month salary , male not receive one month salary) +P( male not receive one month salary , female receive one month salary

= (0.26 x (47/99)x2 ) + ( (0.17x (10/99) x2 ) = 0.28

but the ans is 0.604
2. Relevant equations

3. The attempt at a solution

2. Jul 25, 2014

### HallsofIvy

Staff Emeritus
You are making the same mistake you did in the almost identical problem you posted with "RM" rather than "$". You are using "100" as the denominator when you are given that one person is a male and the other female. Use "36" as the denominator for the male and 64 for the female worker. The probability that a male worker earns less than$1000 is 10/36 and over $1000 is 26/36. The probability that a female worker earns less than$1000 is 17/64 and over \$1000 is 47/64.

3. Jul 25, 2014

### Ray Vickson

There are two possible interpretations:
(1) One male chosen at random from the male workers and one female chosen at random from the female workers.
(2) Two workers chosen randomly; one of them happens to be male and one female.

I suspect that (1) is the intended interpretation, but your working seems to be for (2). How would you tackle (1)?

Note: the two interpretations could give very different answers.