Probability Question

[jabeck74]

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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?

 

[SMC2UCLA2_]

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.

 

[SMC2UCLA2_]

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.

 

[n2o*]

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?

 

[beannaithe]

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.

 

[SMC2UCLA2_]

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.

 

[joooj86]

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

 

[elinz]

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