Problem Solving

Stuart Kovinsky wrote:

Kaunteya wrote: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?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Very similiar to the last problem that I posted but still haven't figured out how to solve these. If someone can please guide that I would really appreciate it. If you could solve this one, could you please solve the other one I posted titled: Combination and Permutation Manhattan. Thanks again guys.

Kaunteya

If 2/3 are men, we have 10 men and 5 women.

We want to know the probability that at LEAST 2/3 of the people actually selected will be men. In other words, that at least 8 out of the 12 jury members will be men.

There are three scenarios in which this could happen:

8 men and 4 women;

9 men and 3 women; and

10 men and 2 women.

Let's see how many different ways we can make each of these occur.

There are 10 men total, so there are 10C8 different groups of 8 men. There are 5 women, so there are 5C4 different groups of 4 women.

Therefore, scenario 1 has 10C8 * 5C4 = 10!/8!2! * 5!/4!1! = 45 * 5 = 225 possible juries.

For scenario 2, we have 10C9 * 5C3 = 10!/9!1! * 5!/3!2! = 10 * 10 = 100 possible juries.

For scenario 3, we have 10C10 * 5C2 = 10!/10!0! * 5!/2!3! = 1 * 10 = 10 possible juries.

[remember, 0!=1]

Now, since this is a probability question, we we want to use the probability formula.

Probability = #desired outcomes / total # of possible outcomes.

We've already calculated the # of desired outcomes: 225 + 100 + 10 = 335 juries with at least 8 men on them.

The total # of possible outcomes is the total # of possible juries, which is simply 15C12 = 15!/12!3! = 15*14*13/3*2*1 = 5*7*13 = lots, so let's reduce instead!

So:

335/5*7*13 = 67/7*13 = 67/91

choose (d).

Let's also look at this question from a strategic guessing point of view.

2/3 of the jury pool are men. Let's eliminate (c) 2/3, because that's way too easy.

Now we have a big split among the remaining choices. (a) and (b) are both very small (less than 1/3) and (d) and (e) are both big (more than 2/3). Since 2/3 of the jury pool are men, does it make any sense that there would be a small probability that 2/3 of the actual jury will be men too? Of course not, so (a) and (b) don't really make sense. So, if we're guessing, choose (d) or (e). (Further, we predict that the answer should be a bit more than 2/3 - (e) really seems too big, so (d) looks like the best guess.)

Never discount the power of strategic guessing, especially on really time consuming questions!

Kaunteya wrote: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?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Very similiar to the last problem that I posted but still haven't figured out how to solve these. If someone can please guide that I would really appreciate it. If you could solve this one, could you please solve the other one I posted titled: Combination and Permutation Manhattan. Thanks again guys.

Kaunteya

Mathsbuddy wrote: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?
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91

Jury:
2/3 of 12 = 8 men
1/3 of 12 = 4 women

Jury pool:
2/3 of 15 = 10 men
1/3 of 15 = 5 women

Therefore success requires that 2 men and 1 women are not selected.

So to choose these 3 people instead would create an equivalent probability (to that required by the question).

Options:
p(MMF)= 10/15 * 9/14 * 5/13 = 15/91
p(MFM)= 15/91
p(FMM)= 15/91

p(A or B or C) = p(A) + p(B) + p(B) = 3 * 15/91 = 45/91

Can anyone identify the flaw in this logic?