Combinations

I think the right question is :

If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?

Solution for this one is

12 people jury out of 15 so the total ways to select a jury = 15C12 = 91 * 5

2/3 men of 15 = 10
women = 5

2/3 of the jury is to be men = 8 men in the jury

question asks about atleast 8 men in the jury
so we will consider less then 8 that is 7 in order to get atleast 8 men on the jury

cases of having 7 men on the jury = 10C7 = 120

probability of having 7 men on jury = 120/(5 * 91) = 24/91 ---- (p7)

probability to have 8 men atleast = 1 - p7
1- 24/91 = 67/91

Could have you done this without using the 7 and using the 8 men instead and not do 1-p7

With the solution with that 1-p(7)

you can only use this because you know 7 men is the only negative outcome. if you were choosing from 16 jurors then you could also have 6 men

Rahul@gurome wrote:

Jinglander wrote:Hey guys sorry I dont have an answer for this

A jury of 12 is to be selected from a pool of 15. 2/3 are male 1/3 are female. What is the probability that the selected jury will be 2/3 male

Here's another way to approach the question:

A jury of 12 members is to be selected, in which 2/3 of the members should be male.
Now 2/3 of 12 = 2*4 = 8. So, the male members can be 8, 9 or 10.
If there are 8 males, then females = 4
If there are 9 males, then females = 3
If there are 10 males, then females = 2

So, No. of favorable possibilities = 10C8 * 5C4 + 10C9 * 5C3 + 10C10 * 5C2 = 45*5 + 10*10 + 1*10 = 225 + 100 + 10 = 335
Total no. of possibilities = 15C12 = 455

So, required probability = 335/455 = 67/91

Does that help?

thank u..is there any way to solve this problem....