Hi Gmatprep550,

The question asks for the PROBABILITY that exactly 2 of the 4 children are boys, so the total number of possible outcomes IS a factor in this question.

If you describe the 5 possible outcomes as...
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl

...then you might think that there is a 1/5 chance of having 2 boys and 2 girls, but that is NOT mathematically correct (since the 5 possible outcomes do NOT all have an equal likelihood of occurring).

GMAT assassins aren't born, they're made,
Rich
_________________

Thanks EMPOWERgmatRichC

Why is it a permutation problem? Why does the sequence matter here?

If it’s asking for a possibility where there are exactly 2 boys and 2 girls.

BBGG
If the sequence doesn’t matter, it’s just 1

Out of

BBGG
BGGG
GGGG
BBBB
GBBB

I got 1/5.

Posted from my mobile device

Hi Seriousthistime,

I discussed this issue in a series of posts directly above yours. The question asks for the PROBABILITY that exactly 2 of the 4 children are boys, so the total number of possible outcomes IS a factor in this question.

If you describe the 5 possible outcomes as...
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl

...then you might think that there is a 1/5 chance of having 2 boys and 2 girls, but that is NOT mathematically correct (since the 5 possible outcomes do NOT all have an equal likelihood of occurring). For example, with 4 children - and an equal probability of having a boy or a girl - there are 2^4 = 16 possible outcomes. The "all girls" and "all boys" outcomes occur just ONCE each (out of those 16). The "two boys and two girls" outcome occurs 6 times.

GMAT assassins aren't born, they're made,
Rich
_________________

without doing the math, can i guess like the following? Bunuel

since the probability of having a boy or girl is equal, then out of 4 children the probability of having 2 boys and 2 girls will be close to 50% or 1/2. only answer choice (A) is close to 1/2. So I will select (A).

BelalHossain046
_________________

Hi RashedVai,

Depending on how the answer choices are written, that type of thinking can potentially be problematic. For example, if the answer choices had included "1/2" as an option....

(A) 1/2
(B) 3/8
(C) 1/4
(D) 3/16
(E) 1/8

... would you have chosen Answer A or would you have done enough work to determine that the correct answer was actually Answer B? While it's a subtle issue, the design of the answer choices can significantly impact how difficult a question actually is to solve - meaning that you have to be careful about the 'assumptions' that you bring into the solving process. By extension, the moment you choose not to do work on your pad to answer a question, then one of 2 results is likely: your thinking is correct OR you're potentially making a silly/little mistake - and you'll never realize it, so you'll get the question wrong.

GMAT assassins aren't born, they're made,
Rich
_________________

Given:A couple decides to have 4 children.

Asked: If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

Total number of ways to have 4 children = 2*2*2*2 = 2^4 = 16

Number of ways to have exactly 2 girls and 2 boys = 4C2*2C1 = 6

The probability that they will have exactly 2 girls and 2 boys = 6/16 = 3/8

IMO A
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for 4 children getting 2 boys and 2 girls ; 4!/2!*2! = 6
and total ways/ chances of getting boys and girls ; 2^4 ; 16
P = 6/16 ; 3/8
OPTION A

I'm always making the same mistake in these types of problems:
I thought we need two G and two B. So 1/2 * 1/2 * 1/2 * 1/2 = 1/16.
But the issue with that is it implies an order such as BBGG. So I need to multiply it by the number of ways we can arrange BBGG which is 5C2.

My question is in what type of problem is my approach correct? When is it correct to multiply the probabilities of the desired result? I remember there was subset of problems in high school where that approach is valid.

When order does not matter use combination

Desired Outcome=!4/(!2*!2)=6
Total Outcome=2^4=16 ( same as Coin, head /tail, boy/girl)
Required Probability=6/16
=3/8
A:)

Hi newdimension,

The calculation that you listed is one that you could use if you were looking for a SPECIFIC outcome. For example, Girl-Girl-Boy-Boy. When you're not concerned about what 'order' the outcome appears - as is the case in this prompt (any combination of 2 boys and 2 girls is what we are after), then you need to either list out all of the options (re: all of the specific orders that fit what you're looking for) or use some variation of the Combination Formula.

GMAT assassins aren't born, they're made,
Rich
_________________

binomial distribution (because the order in which they get a the boy and girl does not matter)

n!/(n-k)! * 1/k! * probability to have a girl^k * probability to have a boy^(n-k)

--> 4!/(4-2)! * 1/2! * 1/2^2 * 1/2^2 = 4*3/2 * 1/4 * 1/4 = 3/8 A