probability
A committee of 5 members is chosen randomly from a group of 4 men and 5 women. What is the probability that the chosen committee contains exactly 3 men and 2 women?
I'm not the best a probability, and learned this method relatively recently, so forgive me if I get this wrong.
Since we want 3 men from a group of 4 men, it would be (4!)/((4-3)!) = (4*3*2*1)/(1!) = (4*3*2) = 24 possible permutations of 3 men. However, the order of these men doesn't matter. A 3-man group of (Man1, Man3, Man2) is the same as (Man2, Man1, Man3). How many ways can we organize a 3-man group? 3 factorial of course, which would be 3*2*1 = 6. This means each 3-man group combination is counted 6 times. In order to find each unique combination, we divide the 24 possible permutations by 6 to get 4 unique combinations of a 3-man group.
In short, this explanation can be simplified to (4!)/(((4-3)!)*(3!)) = (4*3*2*1)/((1)*(3*2*1)) = 4. Notice that this is just 'n' choose 'k', where 'n' is 4 and 'k; is 3.
The logic is the same for women. We choose 2 women from a group of 5 women. (5!)/((5-2)!) = (5!)/(3!) = (5*4*3*2*1)/(3*2*1) = 5*4 = 20 possible permutations of 2 women. A 2 women group of (Woman2, Woman5) is the same as (Woman5, Woman2) so we are counting each group twice. If we only want to count each group once, therefore making unique combinations, divide 20 by 2 to get 10 unique combinations of 2-women groups.
Simplified to (5!)/(((5-2)!)*(2!)) = (5*4*3*2*1)/((3*2*1)*(2*1)) = 10. Again, this is 'n' choose 'k', where 'n' is 5 and 'k' is 2.
Now, we know there are 4 unique combinations of a 3-man group, and 10 unique combinations of a 2-women group. To find the total number of 3-men and 2-women committees, we multiply 4 and 10 to get 40 unique combinations of exactly 3-men and 2-women committees.
We have found the number of committees we can create that are 3-men and 2-women, but not the probability. In order to find the probability, we must find the total number of unique committees possible. There are 9 people in total to choose from, and 5 slots in the committee to fill. Therefore, the number of possible unique committeees is 9 choose 5, which is (9!)/(((9-5!)*(5!)) = (9*8*7*6*5*4*3*2*1)/((4*3*2*1)*(5*4*3*2*1)) = 126 unique combinations of committees.
Of those 126 committees, 40 of them are composed of 3 men and 2 women. Divide 40 by 126: 40/126 = 20/63. Rounding to 4 decimal places we get:
0.3175 * 100% = 31.75% chance that the chosen committee contains exactly 3 men and 2 women.
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