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
Solution:
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 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 is 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.)
a. 24/91
b. 5/91
c. 2/3
d. 67/91
e. 84/91
Solution:
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 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 is 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.)
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