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a family has 5 children, what is the probability that there are 3 boys  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 56% (01:35) correct 44% (02:00) wrong based on 33 sessions

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a family has 5 children, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

Source: www.GMATinsight.com

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Re: a family has 5 children, what is the probability that there are 3 boys  [#permalink]

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1
Probability of a girl or a boy = 1/2
Total children = 5
Probability = 1/2*1/2*1/2*1/2*1/2----1

Now Seq is BBBGG.

So it can be arranged in 5!/3!*2! Ways. ----2

Taking 1 and 2 together

1/32 * 5!/3!*2! = 10/32 = 5/16

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Posts: 7752
Re: a family has 5 children, what is the probability that there are 3 boys  [#permalink]

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GMATinsight wrote:
a family has 5 children, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

Source: http://www.GMATinsight.com

each child could be in two ways - B or G so two ways
total ways of the 5 child = $$2*2*2*2*2 = 2^5=32$$

ways 3 out of 5 are B = $$5C3=\frac{5!}{3!2!}=10$$

probability =$$\frac{10}{32}=\frac{5}{16}$$

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a family has 5 children, what is the probability that there are 3 boys  [#permalink]

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GMATinsight wrote:
a family has 5 children, what is the probability that there are 3 boys and 2 girls among the children?

A) 1/32
B) 1/16
C) 3/32
D) 1/4
E) 5/16

Source: http://www.GMATinsight.com

$$? = P\left( {3B\,\,{\text{and}}\,\,2G\,\,{\text{among}}\,\,5\,\,{\text{children}}} \right)$$

First Step: evaluate the probability of one "typical" favorable scenario,
> Say BBBGG :: 1/(2^5) = 1/32

Second Step: check whether all favorable scenarios are EQUIPROBABLE.
> They are: BBGBG, ... , GGBBB all have this same probability (each one considered separately, of course.)

Third Step: check how many favorable scenarios are there.
> There are C(5,3) = 10 scenarios, because we have to choose 3 places (among 5) to "put" the letter B.
Note that C(5,3) = C(5,2) because choosing 3 places (among 5) to "put" B is equivalent to choosing 2 places to "put" G.

Four Step: check if all scenarios of the previous step are MUTUALLY EXCLUSIVE, so that we can ADD their probabilities next!
> Yes. When (say) BBBGG occurs, BGBBG does not occur... the same applies to any 2 occurrences among the 10 of the Third Step!

Fifth Step: taking into account everything previously considered,
$$? = C\left( {5,3} \right) \cdot \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net a family has 5 children, what is the probability that there are 3 boys   [#permalink] 11 Sep 2018, 06:03
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