Probability of receiving bonus

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SUMMARY

The discussion revolves around calculating the probability of receiving a bonus in a factory with 36 male and 64 female workers. Specifically, 10 male workers earn less than $1000, while 17 female workers earn at least $1000. The correct probability that exactly one worker receives one month's salary is 0.604, contrasting with the incorrect calculation of 0.28. The error stems from using the wrong denominators in the probability calculations, specifically confusing the total number of workers with the specific groups of male and female workers.

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desmond iking
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Homework Statement


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

Homework Equations





The Attempt at a Solution

 
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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.
 
desmond iking said:

Homework Statement


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

Homework Equations





The Attempt at a Solution


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.
 

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