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

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Math Expert
Joined: 02 Sep 2009
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A couple decides to have 4 children. If they succeed in havi [#permalink]

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13 Mar 2014, 02:29
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Difficulty:

75% (hard)

Question Stats:

50% (01:15) correct 50% (01:07) wrong based on 907 sessions

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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: 700

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

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13 Mar 2014, 02:29
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SOLUTION

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 $$\frac{4!}{2!2!}$$.

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

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14 Mar 2014, 04:48
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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Posts: 44419
Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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15 Mar 2014, 10:59
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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25 Apr 2014, 01:06
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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;

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

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14 Mar 2015, 01:30
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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 15 Mar 2015, 00:07, edited 1 time in total.
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Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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14 Mar 2015, 08:14
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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

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

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03 Nov 2015, 21:17
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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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04 Nov 2015, 01:08
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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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Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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23 Feb 2018, 13:22
Hi All,

There are a couple of ways to answer this question:

1) You could make a table of all the options (there are only 16 possible outcomes) and count up all the ways to get 2 boys and 2 girls

or

2) You can do the math

Here's the math approach:

Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities

It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...

4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls

[Reveal] Spoiler:
6/16 = 3/8 = A

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

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18 Mar 2018, 11:27
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide.

I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong.

Could you explain how you determined it to be 600 level difficulty?

Thanks, Bunuel! You're the best
Math Expert
Joined: 02 Sep 2009
Posts: 44419
Re: A couple decides to have 4 children. If they succeed in havi [#permalink]

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18 Mar 2018, 12:07
destinyawaits wrote:
How do you arrive at this being a "600 level difficulty" problem? It's the 8th from the last problem in the OG 18 Supplemental Quant Guide.

I believe it'd be considered at least a 700 level difficulty by GMAC, but I could be wrong.

Could you explain how you determined it to be 600 level difficulty?

Thanks, Bunuel! You're the best

You can check difficulty level of a question along with the stats on it in the first post. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. For this particular question it's 700-level.
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Re: A couple decides to have 4 children. If they succeed in havi   [#permalink] 18 Mar 2018, 12:07
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