Question

Can any One Help This Way Forward?
A population consists of 50% men and 50% women of a population of 50 people. A simple random sample (draws at random without replacement) of 4 people is chosen. Find the chance that in the sample i) there are more women than men? ii) the fourth person is a woman? iii) The third person is a woman, given that the first person and fourth person are both men?

Answer 1

i) there are more women than men?
This means there's either 3 women or 4 women.

So from 25 women, pick 3, and from 25 men pick one, divided by from 50 people pick 4. Add this to the case where you pick all four women (and zero men).

P(3 women or 4 women) = P(3 women) + P(4 women) =
\[\large \frac{ 25C3 \times 25C1 }{ 50C4 } + \frac{ 25C4 }{ 50C4 }\]

Answer 2

ii) the fourth person is a woman?
hmm, this event is dependent on what happens before it - if zero, one, two, or three women are picked before the fourth person. We might need to add up all four cases.

Answer 3

Agentosmith the way forward is not clear and not correct

Answer 4

Did you actually calculate it...? What did you get?

Answer 5

I think I explained it pretty clearly, but I'm not going to just post the answer, you tell me what you got and I'll tell you if it's correct. I'm assuming you know how to calculate combinations.

Answer 6

the answer was 0.0650 and its wrong! My explanation is as follow: You can do it:
You can’t simply look at sets of 4 people, because the order in which they’re chosen matters. Still, you can start by looking at sets and then count the number of usable orderings of each set. I’ll do this below, but note that there is a much easier way to solve this particular problem: since the numbers of men and women in the pool are equal, the fourth person chosen is equally likely to be a man or a woman, and the probability that it’s a woman is therefore ... what?
To calculate the probability by brute force:
• There are (25C4) sets consisting entirely of men; each of these can be drawn in 4! different orders, and none of these 4!⋅(25C4) draws will give you a woman on the fourth draw.
• There are (25C3)(25C1) sets consisting of 3 men and one woman. If the woman is the fourth person drawn, the 3 men can be drawn in any of 3! orders, so there are 3!⋅(25C3)(25C1) ways to draw three men and then a woman.
• There are (25C2)(25C2) sets consisting of 2 men and 2 women. There are 2 ways to choose which woman is drawn fourth. Once you’ve done that, there are 3 ways to decide in which of the first 3 positions the other woman will be drawn, and then there are 2 ways to order the 2 men in the remaining 2 positions. Thus, there are 2⋅3⋅2⋅(25C2)(25C2) ways to draw 2 men and 2 women with a woman in the fourth slot.
• There are (25C1)(25C3) sets consisting of one man and 3 women. The man can be drawn in any of the first 3 slots, and the 3 women can then be drawn in 3! different orders, so there are 3⋅3!⋅(25C1)(25C3) ways to draw one man and 3 women with a woman in the fourth slot.
• Finally, there are (25C4) sets consisting entirely of women, each of which can be arranged in 4! ways to put a woman in the fourth slot, for a total of 4!⋅(25C4) draws of 4 women.
Thus, there are altogether
3!⋅(25C3)⋅25+12(25C2)^2+18⋅25⋅(25C3)+24(25C4)
draws with a woman in the fourth slot, out of a total of 4!⋅(50C4) draws.
________________________________________
Now see if you can do part (ii); it’s actually quite a bit easier. Note that if the first and fourth persons drawn are men, the second and third persons are really being chosen from a pool of 25 women and 23 men. You can now use brute force very easily, though you can also use the idea of the very short solution to (i) and iii to avoid even that.

Answer 7

0.0650 doesn't sound close to reasonable - think about it: there's a 50% chance of picking a woman, so you'd expect to often get about 2 men and 2 women in your group of 4, right?

Occasionally you'll get three women and 1 man, or three men and 1 women; the chance of getting 3 women and 1 man would be higher than 6.5%, and I can be pretty confident of that w/o needing to calculate anything.

for part (i)... The order they're chosen does NOT matter... you're choosing people, order isn't important.

Answer 8

Btw my answer above was not 0.0650 if that's what you meant...

\[\large \frac{ 25C3 \times 25C1 }{ 50C4 } + \frac{ 25C4 }{ 50C4 } = 0.305\]

http://www.wolframalpha.com/input/?i=%2825+choose+3%29%2825+choose1%29%2F%2850+choose+4%29+%2B+%2825+choose+4%29%2F%2850+choose+4%29

Answer 9

@Grad2013 how did you get 0.065...?

And btw in the future please actually make sure you've calculated something correctly before calling someone wrong... 0.065 is not close to the result of the calculation i posted above.

Answer 10

Here my Solution:The Fourth Person is Women Answer is here calculate and get me back=
P(4th Woman)=

=((25C3)/(50C3))*(22/47)+(3C1)*(((25C2)*(25C1))/((50C2)*(48C1)))*(23/47)+
(3C2)*(((25C1)*(25C2))/((50C1)(49C2)))*(24/47)+((25C3)/(50C3))*(25/47)=

Answer 11

again.... @Grad2013 how did you get 0.065...? From what I posted above.

Answer 12

do not get offended if i said its wrong implied my punching came that answer I could be wrong its subjective! but my calculation came 50% different (2300+24)/(230300)+(12650/230300)= 0.065? different 0.305

Answer 13

I even gave You best response then!

Answer 14

Lets work the solution of this 4 part which you solved second and finished! I need your input its a good problem!

Answer 15

P(3rd Woman|1st and 4th Men)=
P(3rd Woman and 1st and 4th Men)/(P(1st and 4th Men))=
((25*24*25*23+25*25*24*24))/((25*24*23)+(22)+(2C1)*(25*24*25*23)+25*25*24*24))= Please calculate and get back to me the answer!

Answer 16

0.065 is not 50% different from 0.305. It's off by a factor of almost five... It's a probability of 31% vs a probability of 7%... you don't see a huge difference there?

(2300+24)/(230300)+(12650/230300) I don't know how you got this, where is 2300+24 coming from...?

\[\large \frac{ 25C3 \times 25C1 }{ 50C4 } + \frac{ 25C4 }{ 50C4 } =\]
\[\Large \frac{ 2300 \times 25 + 12650 }{ 230300 }\]

Answer 17

all 4 parts of this long one which the second one You got the Solution Thanks (2300×25+12650/(230300) I have put in my notes to make sure I punch right Your answer 0.305 was ok but that part I missed my self! Thanks!

Answer 18

P(3 women or 4 women) = P(3 women) + P(4 women) =
25C3×25C150C4+25C450C4
I gave You best response this part a medal! Thanks for your Prompt help then it motivated to solve all other 2 last parts which very very long!