Probability of choosing 4 person

[desmond iking]

Homework Statement

for part b , my working is (12c2 x 18c2)/(35c4) = 0.192

thers're 12 women doctor , 2 are chosen , there're 18 male left, 2 are chosen . no condition= choose 4 person from 35 person.

but the ans is 0.807

Homework Equations

The Attempt at a Solution

 

[Orodruin]

In the case of b the men have nothing to do with it. You already know that two women were selected and you must condition your probability to this, i.e., given two women from the attendees - what is the probability that both are doctors?

 

[D H]

Both your answer and the answer sheet are wrong with regard to part b.

You computed the probability that a randomly selected panel will have two female doctors. That is not the question. Some of those randomly selected panels will have zero females, others one, others three, yet others four. You need to exclude all of these because it's a given that two of the panel members are female.

I don't know what the answer sheet calculated, but it too is wrong.

 

[desmond iking]

In the case of b the men have nothing to do with it. You already know that two women were selected and you must condition your probability to this, i.e., given two women from the attendees - what is the probability that both are doctors?

why the men have nothing to do with the 2 women doctor chosen? the question want 4 members in the panel. am i right? if so, my working is 12x11 /(17x16)=0.49

since we already know there're 17 women , after 1 is chosen 16 left. there're 12 woman doctor, after 1 is chosen, 11 left. am i correct?

 

[HallsofIvy]

why the men have nothing to do with the 2 women doctor chosen? the question want 4 members in the panel. am i right? if so, my working is 12x11 /(17x16)=0.49

since we already know there're 17 women , after 1 is chosen 16 left. there're 12 woman doctor, after 1 is chosen, 11 left. am i correct?

Part b says "given that two women are selected what is the probability that both are doctors". That is exactly the same as if the problem had said "if there are two women chosen from 17, 12 of whom are doctors, what is the probability both women chosen are doctors?" without mentioning men at all. Yes, that is \left(\frac{12}{17}\right)\left(\frac{11}{16}\right).

 

[desmond iking]

Part b says "given that two women are selected what is the probability that both are doctors". That is exactly the same as if the problem had said "if there are two women chosen from 17, 12 of whom are doctors, what is the probability both women chosen are doctors?" without mentioning men at all. Yes, that is \left(\frac{12}{17}\right)\left(\frac{11}{16}\right).

for part a , 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


the correct working would be (20C3 X15C1)/35C4 = 0.327

Is my concept wrong? D=doctor E=engineer

why by using the method of C (combinations) , the arrangment of person is not importamt , which means i don't need to times 4 compared to the method i used above

 

[desmond iking]

Can anyone reply,please? Thanks in advance!

 

[Orodruin]

You can do it in either of the ways. In the first way, you are taking the probability of choosing a particular ordered set and there are 4 such ordered sets that give you the sought result.

When you study the number of combinations you are studying the number of unordered sets both in denominator and numerator.

 

[haruspex]

I suspect part (b) is intended to be interpreted in the context of part (a), i.e., given 3 doctors are selected and two women are selected. This gives an answer much closer to the book one, but I still don't get .807 (which is very close to 46/57). I get 22/29.

 

[D H]

Hmm. That's a possibility. I get 38/45 with that interpretation. (Again, not 0.807.)