Probability Q

[punitkaur]

. What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) 1/4
c) 1/2
d) 3/8
e) 5/8

My answer is 1/8 but the answer given is 3/8

[jzdchou]

It is like flipping a coin 3 times and getting 8 different total outcomes.

1/2 * 1/2 * 1/2 = 1/8

Then I just list the possibilities
BBB
BGG
BBG
GBB
GGB
GBG
BGG
GGG

2G and 1B = GGB, GBG, BGG = 3/8

[punitkaur]

Thanks jzdchou. I did a silly mistake while solving the problem, i took all 3 cases GGB, GBG, BGG to be the same and hence got 1.

Did not realize that sequence does matter.

[jzdchou]

Or once you get total of 8 combinations....

instead of doing the listing, think of how many ways can 2G and 1B be arranged/ordered.

Isn't it simply 3!/2! = 3*2*1/2*1 = 3?

[crackthetest]

How to know if you need to use the permutation here, so that order really matters. I see the question just asks for one boy & 2 girls but not how they are arranged.

Pl. clarify.

[Stuart@KaplanGMAT]

How to know if you need to use the permutation here, so that order really matters. I see the question just asks for one boy & 2 girls but not how they are arranged.

Pl. clarify.

Since the question doesn't mention a specific order, all orders are acceptable. Since all orders are acceptable, we have to take them all into account.

Identifying this as a pseudo-coin flip question makes life much easier, especially if you know the coin flip formula:

Prob(k results out of n flips) = (nCk)/2^n

In this question, we want 2 boys out of 3 kids, so let's let n=3 and k=2:

(3!/2!1!)/(2^3) = 3/8

Note that we could have looked at the question from the other side: we want 1 girl out of 3 kids, so let's let n=3 and k=1:

(3!/1!2!)/(2^3) = 3/8

Any question with a 50/50 probability can be solved using the above formula.

For more info about coin flip and pseudo-coin flip questions, check out this thread:

https://www.beatthegmat.com/coin-flip-qu ... 17911.html

[crackthetest]

Thanks very much Stuart, your link to pseudo-coin is very helpful.