A couple decides to have 4 children. If they succeed in having 4 child

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

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.

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

Hi EMPOWERgmatRichC generis chetan2u VeritasKarishma

How can I determine that permutations are required here ? Any thumb rule ?

The probability of getting 2 boys and 2 girls in any order is 1/16. By the wording of the question it appears that the question is asking for the probability of any one of these permutations of BBGG (as the end result will be same). Why are we considering BBGG to be different from BGBG ?

Please help to get a hang of this concept.

Hi altairahmad,

Most GMAT questions can be approached in more than one way. For a question that is specifically about a Permutation, then you should look for a reference to the 'order' of an outcome (for example, the number of different ways for 1st-2nd-3rd to be won in a race).

You can approach this question either with Permutations OR with the Combination formula. If you're going to use Permutations here, then that work requires that you have to essentially "list out" every way for 2 boys and 2 girls to occur (thankfully, there aren't actually that many ways) AND understand that there are only 16 possible outcomes.

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

Thanks for the reply.

I am still having trouble understanding that why the answer is not simply 1/16 i.e probability of having 2 boys and 2 girls. Why do we have to consider order.

1st Child can be a Boy or a Girl = 2 options. Similarly, for the remaining 3 we will have 2 options each.

Total outcomes: \(2^4\) = 16.

Desired: Exactly 2 girls and 2 boys: GGBB

This 4 letters can be arranged in 4! ways.

But as the Girls are identical, and so are boys.

=> Desired result: \(\frac{4! }{ 2! 2!}\) = \(\frac{24 }{ 4}\) = 6

Probability : \(\frac{Desired }{ Total }\)

=> \(\frac{6 }{ 16 }\)

=> \(\frac{3 }{ 8 }\)

Answer A

Hi altairahmad,

You understand that there are 16 possible 'arrangements' of the 4 children (since there is an equal chance for any child to be a boy or a girl, the total number of arrangements would be 2x2x2x2 = 16)... so it might help to consider the order in which the children are born (and you can refer to them as 1st, 2nd, 3rd and 4th).

Consider the various ways that you could end up with 2 boys and 2 girls:

It could be BBGG (meaning that the first two children were boys and the third and fourth were girls). IF that was the only possibility, then the answer would be 1/16. However, that is NOT the only possibility; there are others (including, BGBG or GGBB). It shouldn't take you long to figure them out - and once you do, you'll have the answer to the question.

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

What are the total ways of having 4 children?

The first child in 2 ways, second in 2 ways, third in 2 ways and fourth in 2 ways to give 2^4 = 16 ways

This will include all cases: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, BBGG, GGBB, GBGB, GBBG, BGGB, GGGB, GGBG, GBGG, BGGG, GGGG etc

We need those which has 2 boys and 2 girls. How many such cases are there?
BGBG, BBGG, GGBB, GBGB, GBBG, BGGB (every way in which 2 boys and 2 girls are arranged is acceptable)

Since arrangement doesn't matter, we need to accept every arrangement (and that is the reason you need to find the number of arrangements)

Probability = 6/16 = Acceptable cases/Total number of cases

Solution :

The total possible sample space or outcome = 2^4 = 16

Total favorable ways = 6
GBGB
GGBB
BBGG
BGBG
GBBG
BGGB

Alternately, the favorable outcomes = 4!/ (2!x2!) =6

Thus the probability = 6/16 = 3/8 (option a)

Devmitra Sen
GMAT SME
_________________

We can solve this question using counting methods.
P(exactly 2 girls and 2 boys) = (number of 4-baby outcomes with exactly 2 girls and 2 boys)/(TOTAL number of 4-baby outcomes)

As always, we'll begin with the denominator.

TOTAL number of 4-baby arrangements
There are 2 ways to have the first baby (boy or girl)
There are 2 ways to have the second baby (boy or girl)
There are 2 ways to have the third baby (boy or girl)
There are 2 ways to have the fourth baby (boy or girl)
By the Fundamental Counting Principle (FCP), the total number of 4-baby arrangements = (2)(2)(2)(2) = 16


Number of 4-baby outcomes with exactly 2 girls and 2 boys
This portion of the question boils down to "In how many different ways can we arrange 2 G's and 2 B's (where each G represents a girl, and each B represents a boy)?"

----------ASIDE-------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
------------BACK TO THE QUESTION---------------------------

Our goal is to arrange the letters G, G, B, and B
There are 4 letters in total
There are 2 identical G's
There are 2 identical B's
So, the total number of possible arrangements = 4!/[(2!)(2!)] = 6


So.....
P(exactly 2 girls and 2 boys) = 6/16 = 3/8

Answer: A

Hi KarishmaB.

I am still confused.. It is being stressed that order matters here. But in the question, it seems, only the count matters.

.

Think about it - the probability that a boy is born is 50% and girl is born is 50%.
So if one decides to have 4 kids, what is most likely? That half will be boys and half girls, right? What will be the probability that say all are boys? Will it be same as that of 2 boys and 2 girls? No, it will be much less likely. Hence 1/5 is not correct .
Of the 16 ways in which we can have 4 kids, only 1 way has all boys. 1 way has all girls. In 4 ways, we will get 3 boys and 1 girl and in 4 ways we will get 3 girls and 1 boy.
And in 6 of those cases, we will have 2 boys and 2 girls. So probability will be 6/16.

To get this value of 6, we arrange to see. Since arrangements are irrelevant, we take all 6 to be our possible cases but we must find that there are 6 such cases.

Posted from my mobile device

Could please explain that MISSISSIPPI rule?

That rule just reflects the treatment of arranging when there are multiple instances of the same object, for example MISSISSIPPI containing multiple S and P, etc.

So, for example, arranging 5 different objects is a permutation, calculated by

5!= 24

But if 2 of the 5 are actually the same, then need to adjust by dividing by 2!.

This process is highlighted in many of the previous responses.

Posted from my mobile device

Thank you !

So if there is 8 letters (3 for one, and 5 for the other). You want the number of different ways to arrange.
It would be 8!/5!3! ?

Same way for 5 letters (3 and 2)
It’s 5!/3!2!

Am I right for those 2 examples ?

You only divide by the number of objects that are identical, so if it's

ABCDD

then it's 5!/2!

But if it's

AAABB

then it's

5!/2!3!

Posted from my mobile device