# Thread: probability of receiving bonus

1. ## probability of receiving bonus

A factory of 36 male workers and 64 female workers. worth 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. ## Re: probability of receiving bonus

here's another part of this question,

find the probability of the three doctors selected . the correct working would be (20C3 X15C1)/35C4 = 0.327

can i do in this way? P(DDDE) + P(EDDD) + P(DEDD) +P(DDED) =
( (20/35) x (19/34) x (18/35) x (15/32) ) x 4 = 0.327

Is my concept wrong? D=doctor E=engineer

3. ## Re: probability of receiving bonus

Originally Posted by delso
A factory of 36 male workers and 64 female workers. worth 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
fixed \$signs 4. ## Re: probability of receiving bonus Originally Posted by romsek fixed \$ signs
find the probability of the three doctors selected . the correct working would be (20C3 X15C1)/35C4 = 0.327

can i do in this way? P(DDDE) + P(EDDD) + P(DEDD) +P(DDED) =
( (20/35) x (19/34) x (18/35) x (15/32) ) x 4 = 0.327

Is my concept wrong? D=doctor E=engineer