[snickerskazzoo]

A family is going to have four children. Find the probability for each of the following events.
1. the family has exactly 2 male children
2. the family has at most 2 male children
3. the family has 4 male children

[Kahani Punjab]

Snicker Skazzoo,

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

Here, four kids are in the offing. So, the possibilites are (4M, 0F), (3M, 1F), (2M, 2F), (1M, 3F) and (0M, 4F). They are 5 i.e. FIVE possibilites

Probability of an event = favourable events/Total possible events.

1. Here, favourable events = (2M, 2F), only 1
P (exactly 2 male kids) = 1/5

2. In this case, favourable events = (0M,3F), (1M,3F), (2M, 2F), totalling 3

P (atmost 2 male children) = 3/5

3. In third case, favourable events = (4M,0F) counting 1 only

P (4 male children) = 1/5

This is the answer, which you wanted to know.
I shall feel lucky to help you out, if you have such more questions. Write such probability queries here, and I shall solve them all (if I could) for you, and all other present here, who want such help. Click at HELPFUL, if you find it helping.

[Unknown008]

This problem could be made easier with the binomial distribution, but let's do the usual way.

4. P(4 Males) = P(1st child is male) x P(2nd child is male) x P(3rd child is male) x P(4th child is male)
= 1/2 * 1/2 * 1/2 * 1/2
= 1/16

1. P(2 Males) = Either of:
Male, Male, Female, Female
Male, Female, Male, Female
Male Female, Female, Male
Female, Male, Male, Female
Female, Male, Female, Male
Female, Female, Male, Male

This gives for each, for a total of

At most 2 is when they have 0 male, 1 male or 2 male. You have the probability for 2 males already. Find that of 1 male, and that of 0 male.

Edit: Typo on first probability

[ebaines]

You're all wrong. Sorry - the probability of having 2 males and 2 females is NOT 1/5. It's solved like this:

P(2 males) = (1/2)^4 x C(4,2) = 3/8.

To prove it you can write out all 16 possible combinations of males and females and you'll find that 6 of them involve 2 males. Hence P(2 males) = 6/16 = 3/8.

Similarly:
P(0 males) = (1/2)^4 x C(4,0) = 1/16
P(1 male) = (1/2)^4 C(4,1) = 1/4
P(2 males) = (1/2)^4 C(4,2) = 3/8
P(3 males) = (1/2)^4 C(4,3) = 1/4
P(4 males) = (1/2)^4 C(4,4,) = 1/16.

If you add up all of these outcomes you see the total is 1, as expected.

Kahani and Ambitious: - the problem with your approach is that it assumes incorrectly that all five of the outcomes you wrote are equally probable, but they are not.

Unk - simple math error! (1/2)^4 = 1/16, not 1/8.

[Unknown008]

That's what happens when I multitask at work :p

And my second answer based off on the first, so yea, lol. Sorry about that.