How do you do this?

If a woman has 4 kids, what is the probability that 1 will be a girl and 3 will be boys?

How do you do this?

If a woman has 4 kids, what is the probability that 1 will be a girl and 3 will be boys?

oops wrong, gimme a sec.

oops wrong, gimme a sec.

Ok. here it is.

The probability of a boy is 1/2 which is also the probability of a girl.

There are several ways to get 3 boys and 1 Girl and in this question the order doesnt matter so we are looking at the following combinations: GBBB, BGBB, BBGB, and BBBG (B= prob of boy and G = prob of girl).

You can find the probability of each of these seperately and each will = 1/16. (1/2 x 1/2 x 1/2 x 1/2)

Your final answer will be P(GBBB) + P(BGBB) + P (BBGB) + P(BBBG) = 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4.

OR... you can use Binomial Expansion to determine the probability of a particular combination rather than going through all possibilities.

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

Notice the coefficients for each term come from pascals triangle. In this example we would be looking at (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.
lets let a = prob of girl and b = prob of a boy and a = b = 1/2.

According to the binomial expansion, the probability of 3 boys and 1 girl will be 4a3b where 3 is an exponent.

So we get.... 4(1/2)^3 (1/2) = 4/16 = 1/4.

Ok. here it is.

The probability of a boy is 1/2 which is also the probability of a girl.

There are several ways to get 3 boys and 1 Girl and in this question the order doesnt matter so we are looking at the following combinations: GBBB, BGBB, BBGB, and BBBG (B= prob of boy and G = prob of girl).

You can find the probability of each of these seperately and each will = 1/16. (1/2 x 1/2 x 1/2 x 1/2)

Your final answer will be P(GBBB) + P(BGBB) + P (BBGB) + P(BBBG) = 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4.

OR... you can use Binomial Expansion to determine the probability of a particular combination rather than going through all possibilities.

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

Notice the coefficients for each term come from pascals triangle. In this example we would be looking at (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.
lets let a = prob of girl and b = prob of a boy and a = b = 1/2.

According to the binomial expansion, the probability of 3 boys and 1 girl will be 4a3b where 3 is an exponent.

So we get.... 4(1/2)^3 (1/2) = 4/16 = 1/16.

um, that's too long of an explanation. i don't think the question is meant to be solved that way. here is my suggestion:

the probability of having either a girl or boy is 1/2. so the probability of having one girl AND three girls will be (1/2)^4 = 1/16. it's the same as asking the probability of having one girl AND one boy AND one boy AND one boy.

I think this is right. anyone else?

THe second explination is the easiest one to reason through on the exam. even though the first one has the right math behind it, it might be a little hard to remember during the test.

um, that's too long of an explanation. i don't think the question is meant to be solved that way. here is my suggestion:

the probability of having either a girl or boy is 1/2. so the probability of having one girl AND three girls will be (1/2)^4 = 1/16. it's the same as asking the probability of having one girl AND one boy AND one boy AND one boy.

I think this is right. anyone else?

No, i think this is wrong. I made a typo. If you read my other post you will see that I said the answer is 1/4th in the first answer and then "4/16 = 1/16" in the second method. I obviously didnt do the reduction correctly. The answer is not 1/16. The answer is 1/4th.

it's the same as asking the probability of having one girl AND one boy AND one boy AND one boy.

the probability of two boys will be (1/2)^2 = 1/4
the probability of a girl and a boy will be (1/2)^2 + (1/2)^2 = 1/2

This is because there is only one way to get two boys. Baby #1 is a boy and Baby #2 is a boy.

There are two ways to get 1 boy and 1 girl. BG & GB (in that order).
P(BG) = (1/2) x (1/2) = 1/4
P(GB) = (1/2) x (1/2) = 1/4

The probability of 1 boy and 1 girl will be 1/4 + 1/4 = 1/2
For problems as simple as this its easier to write out all possibilities of births. For problemts with 4+ births you may find it easier to use the binomial expansion.


I recommend you read and understand one of the methods I have outlined in post #3. There is no shortcut. You need to understand the question if you want to get these right.

um, that's too long of an explanation. i don't think the question is meant to be solved that way. here is my suggestion:

the probability of having either a girl or boy is 1/2. so the probability of having one girl AND three girls will be (1/2)^4 = 1/16. it's the same as asking the probability of having one girl AND one boy AND one boy AND one boy.

I think this is right. anyone else?

i dont think its right basically here is the actual equation
(4!/ 1! 3!) * (1/2)^1 (1/2)^3 ..this is an important formula ..will help u a lot

I guess another easy way to do this is:

since you want one girl OR 1 boy somewhere there, that's (1/2)+(1/2)
but then you want two more boys which is (1/2)x(1/2)

Multiply that since that's the arrangement you want, 1 girl and 3 boys,
(2/2)x(1/4)=1/4