A family has 5 children, what is the probability that there are 3 boys

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

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.

\(p(boy or girl)=1/2\)
\(arrangements(BBGGG)=5!/3!2!=10\)
\(p(BBGGG)=(1/2)^5•10=10/32=5/16\)

Answer (E)

GMATinsight help plz

why does it have to be a case of ordering? question didn’t say anything about order...it simply asked if there are 3 boys and 2 girls...

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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: https://www.GMATinsight.com

 
­I have the same question as well. Why should there be ordering? IMO, the answer must be \(\frac{1}{32}\) as this would be logical. I think there must be additional information in the question so that an "ordering" sequence should be initiated?

Bunuel, BrentGMATPrepNow, JeffTargetTestPrep, ScottTargetTestPrep, KarishmaB: could you help here please?­

­
There are 5 children XXXXX, each of them has 2 options, either a boy or a girl, so there are a total of 2*2*2*2*2 = 32 different cases:
BBBBB
BBBBG
BBBGB
BBGBB
BGBBB
GBBBB
BBBGG
...

The cases when there are 3 boys and 2 girls are cases with all arrangements of BBBGG, which is 5!/(3!2!).

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

Answer: BBBGG (can be arranged in 5!/(3!*2!) = 10, 3 boys and 2 girls

Probability: (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)*arrangements i.e 10
answer: 5/16 (E)