A group consists of 5 males and 6 females. Four people are to be selected from the group.
1. in how many ways can we choose 4 people from those 11?
2. In how many ways can we choose 4 females from the 6 females?
3. What is the probability that we get all 4 females when we choose 4 from the whole group?
4.What is the probability that we get at least 1 male when we choose 4 from the whole group?

Snicker Skazzoo,

First, let me do the 3rd question of the above query.

The basic rule to find probability of some event is to divide the favourable events by the total number of possible events.

Here, four persons are chosen from the group of 5m+6f = 11. So the possibilites are (4M, 0F), (3M, 1F), (2M, 2F), (1M, 3F) and (0M, 4F). They are 5 i.e. FIVE possibilities. But we want the possibility – (0M, 4F)
There are 6 females (and 5 males), from which the first can be chosen in 6 ways (i.e. possibility of every girl out of 6 has same chance), so the first girl can be chosen in 6/11 ways, as 11 are total persons. The 2nd can be chosen in 5 ways (i.e. possibility of every girl out of 5 has same chance), so the first girl can be chosen in 5/10 ways, as 10 are remaining total persons. Similarly, the probability for choosing 3rd and 4th girl are 4/9 and 3/7 respectively.

All these happen together, so, we are to multiply all the possibilities, and the answer is = 6/11*5/10*4/9*3/7 = 4/77

This is the answer, which you wanted to know.
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1. and 2. are combinations/permutations problems.

Number of ways to get x number of people from y number of people is given by: yCx, or

From there, you will use those answers for 1 and 2 to get your answer to 3, where the answer to 3 is given by your answer to 2 divided by your answer to 1.

4. This becomes the number of ways where you have not 'all females', or 1-2 to give the number of ways there is at least one male, and then that divided by 1. to get your answer.

PS: Punjab, you did a mistake in your solution.

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