Probability of receiving bonus

[desmond iking]

Homework Statement

please refer to the photo, can i redo the question in this way ?

P( male receive, female not receive) +P( female receive , male not receive)

( (10/1000)x (170/999) ) + ( (470/1000)x (260/999) ) = 0.1393

this is based on 'without replacement' is my concept wrong?

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

You are given that, of 360 male workers, "100 earn less than RM2000.00 a month".
In your calculation for b, you use a probability of .10. That would be 100/10000, the probability that a randomly chosen worker is male and earns less than RM2000.00 a month. But you are told that a male and female worker are chosen so you should not include the probability a worker chosen is male. The probability the male worker earns less than RM2000.00 a month is 100/360, not 100/1000.

 

[desmond iking]

You are given that, of 360 male workers, "100 earn less than RM2000.00 a month".
In your calculation for b, you use a probability of .10. That would be 100/10000, the probability that a randomly chosen worker is male and earns less than RM2000.00 a month. But you are told that a male and female worker are chosen so you should not include the probability a worker chosen is male. The probability the male worker earns less than RM2000.00 a month is 100/360, not 100/1000.

so the ans would be ( (260/350) x (470/640)) + ( (100/360) x (170/640 )) = 0.604 ?

 

[Orodruin]

Yes.

As a curiosity, the problem as stated is not solvable as the workers earning exactly 2000.00 are included in the group earning a month's salary as bonus and thus will also get a 2000.00 bonus. We are not given the number of such workers.

 

[desmond iking]

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

 

[desmond iking]

Yes.

As a curiosity, the problem as stated is not solvable as the workers earning exactly 2000.00 are included in the group earning a month's salary as bonus and thus will also get a 2000.00 bonus. We are not given the number of such workers.

by doing this ( (260/350) x (470/640)) + ( (100/360) x (170/640 )) = 0.604 ,
i assume that P(male receiving bonus, girl not receiving bonus) + P(girl receiving bonus , boy not receiving bonus)

why there's also probability that girls picked first and not receiving bonus , then male receiving bonus is picked after this for P(male receiving bonus, girl not receiving bonus) ? and the same thing goes to P(girl receiving bonus , boy not receiving bonus) ... why the sequence is not important ?


why can't i do in this way? ( (260/350) x (470/640)x2) + ( (100/360) x (170/640 )x2) , but by doing so my ans is more than 1 , which is indeed not correct.