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

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

A couple decides to have 4 children. 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?

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

Problem Solving
Question: 160
Category: Arithmetic Probability
Page: 83
Difficulty: 600


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A couple decides to have 4 children. 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?

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

\(P(GGBB)=\frac{4!}{2!2!}*(\frac{1}{2})^4=\frac{3}{8}\), we should mulitply by \(\frac{4!}{2!2!}\) as the scenario GGBB can occur in several ways: GGBB, BBGG, GBGB, ... # of scenarios possible is # of permutations of 4 letters GGBB out of which 2 G's and 2 B's are identical and equals to \(\frac{4!}{2!2!}\).

Answer: A.
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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New post 14 Mar 2014, 03:48
Would Pascal's Triangle be a useful tool to quickly arrive at the answer in this case?
Using the triangle, I'm getting 3/8 as the answer, which I think is right, but I don't know if there are any limitations / specific cases in which Pascals Triangle should not be used for calculations like these. Please advise

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21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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no of ways of getting P(GGBB) is 4!/2!*2!;
Total no of ways is 2^n =2^4 =16;

6/16 = 3/8;
We can consider this question to a coin that is flipped for 4 times . what is the probability of getting exactly two heads .

P(all out comes) = 1/2 *1/2 *1/2 *1/2 =1/16;

P(favorable outcomes) = 4!/(2! * 2!) = 6/16 =3/8;

(OR)

Second Approach

GBGB
GGBB
BBGG
BGBG
GBBG
BGGB

6 possible ways .
total no of ways is
Baby can be a boy or a girl.
For each baby the probability is 1/2 ;for 4 babies it's 1/16;
6/16 = 3/8;

Press Kudos if this helps :)

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So here we have four positions and there are two options to fill each position
so total number of cases=2x2x2x2=16
now we need 2boys and 2 girls OR we can say that we simply need 2 boys because if its not a boy it has to be a girl
favourable cases=4C2=6
probability=6/16=3/8

(A)
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Last edited by masoomdon on 14 Mar 2015, 23:07, edited 1 time in total.

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1/2 is the chance of boy or girl

1/2 * 1/2 * 1/2 * 1/2 = 1/16 (4 children, boy or girl)

Possible ways of 2 boys, 2 girls:

GGBB
BBGG
GBGB
BGBG
BGGB
GBBG

= 6 ways

we need "OR" --> 1/16 OR 1/16 ....
1/16 + 1/16 + 1/16 ....... +1/16 = 6/16 = 3/8

Answer A

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Binomial Probability
The probability of achieving exactly k successes in n trials is shown below.
Formula: P(Probability of K successes in n trials) = nCk p^k q^n-k

n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial

According to our question

n(4 children) = 4
k( we want exactly 2 girls) = 2
n – k = 2
p (probability of getting a girl in one trial) = 1/2
q = 1 – p = 1/2

4C2 * 1/2^2 * 1/2^2 =3/8

Binomial can also be used for problem like coins.

Binomial is used when following conditions are satisfied.

Fixed number of trials
Independent trials
Two different classifications
Probability of success stays the same for all trials

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Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

A couple decides to have 4 children. 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?

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



Let us handle a single case. First boy, second boy, third girl, fourth girl. The probability of this case is 1/2*1/2*1/2*1/2 = 1/16
Two boys or two girls can be born among 4 children in 4C2 ways= 6 ways
So the required probability is: total number of ways * probability of one such way= 6* (1/16)=3/8
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