probability question?

Question

Two canoe riders must be selected from each of two groups of campers. One group consists of 3 men and 1 woman, and the other group consists of 2 women and 1 man. What is the probability that 2 men and 2 women will be selected?

*pls include details as to how to solve it. thanks!

Answer 1

your question is solvable by:

P = P1 x P2

which means that they are independent probabilities (by groups) that are dependent to each other, hence we have the multiplication

P! = 3/4 or 0.75 probability that two men will be selected from the first group

P2 = 2/3 or 0.667 chance that two women will be selcted from the second group

this gives us

P = 3/4 x 2/3
P = 72/144 or 0.50 or 50% chance that it will ever happen

cheers!!

Answer 2

First write down the number of ways you can get what you want: 2 men and 2 women.

First group: 2 men --- second group: 2 women
First group: 1 man, 1 woman --- second group 1 man, 1 woman

Now for the probabilites of each:
==========================================
2 men from first group:
first group has three men so there are 3 ways. If 1,2 and 3 are the men then (1-2) (1-3) (2-3) and there are 6 total outcomes (these plus selecting the woman and one of the 3 men).
So the probability of getting two men from the first goup is 3/6 or 1/2.

Now to get two women from the second group. Using 1 and 2 as the women, then there is one way: (1-2) out of a total of 3 ways (this plus the woman and each of the men). So the probability is 1 out of 3 or 1/3.

Combine these two to get the probability for 2 men from the first group and 2 women from the second: (1/2)(1/3) or (1/6)

======================================
First group: 1 man, 1 woman. From above this is seen to be 1/2 (3 ways out of the total of 6).

second group 1 man, 1 woman. From above this is seen to be 2/3 (two ways out of the total of 3)

Combine these: (1/2)(2/3) =1/3

========================================
Combine the two computed probabilities: 1/6 + 1/3 = 1/2

So the probability of getting two men and two women is 50 percent

Answer 3

Group A:
prob 2 men = 3/4 * 2/3 = 1/2
prob 1 man 1 woman = 3/4 * 1/3 + 1/4 * 1 = 1/2

Group B:
Prob 2 men = 0
Prob 1 man 1 woman = 1/2 * 2/3 + 1/2 * 2/3 = 2/3
Prob 2 women = 2/3 * 1/2 = 1/3

So prob 2 men and 2 women =
1/2 * 1/3 + 1/2 * 2/3 = 1/6 + 2/6 = 3/6 = 1/2
.

Answer 4

First, let's get some guaranteed things out of the way. In the 1st group, since you are taking 2 of the 4 and there is only 1 woman, 1 of the 2 chosen will ALWAYS be a man. So, I can simplify the 1st group into always removing a man as first choice, which simplifies the 1st group to 2 men and 1 woman. Odds of a woman: 1 of 3. Odds of a man: 2 of 3 Now for group 2: as with group 1, since there are 2 women and 1 man, I am guaranteed having a woman as 1 of the 2 chosen. After removing a woman as 1st choice, the remainder is 1 of 2 for either. Next, since I have removed a man and a woman as guaranteed 1st choices, the second choice resolves down to the following:

1st group: woman=1 of 3, man = 2 of 3
2nd group: woman=1 of 2, man = 1 of 2

There are the following combinations available:
1st group consists of labels: woman1, man1, man2
2nd group consists of labels: woman2, man3

Which makes for exactly 6 combinations

woman1+man3 *
woman1+woman2
man1+woman2 *
man1+man3
man2+woman2 *
man2+man3

The starred combinations meet the criteria, which you can see by inspection is exactly half, which means 50-50. Odds are that if you choose 2 from group 1 and 2 from group 2, your odds of 2 men and 2 women are exactly 50-50, or 1 in 2.

There you go, no messy mathematics, just simplification of the problem because of certain guaranteed outcomes (1 of the 2 chosen from group1 MUST be a man and 1 of the 2 chosen from group2 MUST be a woman), and inspection of the remaining combinations to calculate the odds. It is strange that with an odd number of people, 7, the odds of a 2 men + 2 women as passengers is 50-50.

Answer 5

Group 1 has 3 Males and 1 female and you will select 2 of them.
there are 12 ways to select two people and of these there are three pairs with one female. thus from group 1

P[1 W and 1 M] = 1/4
P[2 M ] = 3/4

Group 2 has 2 W and 1 M. take two and there are 3 ways to do this. Of which there is only one that has two females.

P[2W] = 1/3 and P[1W and 1M] = 2/3

The total probability of having two males and two females is found by summing the probabilites of the two ways this could happen.

P[2M from G1]*P[2W from G2] = 3/4*1/3 = 1/4
P[1M and 1W from G1]*P[1M and 1W from G2] = 1/4*2/3

the sum is 1/4 + 2/12 = 5/12

Answer 6

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