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Family with three children

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Manager
Manager
Joined: 19 Oct 2010
Posts: 242
Location: India
Concentration: Marketing, Finance
Schools: Booth '14, Ross '14, Yale '14, Fuqua '14, Kelley '14
GMAT 1: 560 Q36 V31
GPA: 3
Followers: 5

Kudos [?]: 10 [0], given: 21

GMAT Tests User
Family with three children [#permalink] New post 06 Feb 2011, 02:20
00:00

Question Stats:

33% (00:00) correct 66% (00:00) wrong based on 1 sessions
What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

    A. 1/8
    B. 1/4
    C. 1/2
    D. 3/8
    E. 5/8

Can someone please help with this?
[Reveal] Spoiler: OA

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petrifiedbutstanding

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Joined: 20 Dec 2010
Posts: 2100
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Kudos [?]: 655 [0], given: 376

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Re: Family with three children [#permalink] New post 06 Feb 2011, 02:52
Probability(1B,2G) = Favorable choices of having exactly 1B and 2G/Total number of Possiblities of sexes for 3 children

Let's do the sample set:

Represent Girls with G and Boys with B

3 children can have following possibilities of sexes

GGG - 3 girls
GGB - 2 girls one boy
GBG
GBB
BGG
BGB
BBG
BBB

Total possibilities of both sexes = 8

Number of choices where there are exactly 2 girls and 1 boy are:
GGB
GBG
BGG
=3.

Thus, probability = 3/8

Ans: "D"

This problem is similar to having exactly 2 heads in 3 tosses.

Alternate way;
The total number of possibilities = (Number of possible outcomes in each flip)^(Number of tosses) = 2^3=8

Likewise;
The total number possible sexes for 3 children = (Number of possible sex for each child)^(Number of children)
Number of possible sex for each child = 2 = (Boy or Girl)
Number of children = 3

Total possible sexes = 2^3 = 8

Possibilities to have exactly 2 Girls out of 3 child and 1 Boy out of remaining 1 Child:
=
C^{3}_{2}*C^{1}{1} = 3

Thus, probability = Favorable/total outcomes = 3/8.
_________________

~fluke

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Manager
Manager
Joined: 19 Oct 2010
Posts: 242
Location: India
Concentration: Marketing, Finance
Schools: Booth '14, Ross '14, Yale '14, Fuqua '14, Kelley '14
GMAT 1: 560 Q36 V31
GPA: 3
Followers: 5

Kudos [?]: 10 [0], given: 21

GMAT Tests User
Re: Family with three children [#permalink] New post 06 Feb 2011, 02:59
fluke wrote:
Probability(1B,2G) = Favorable choices of having exactly 1B and 2G/Total number of Possiblities of sexes for 3 children

Let's do the sample set:

Represent Girls with G and Boys with B

3 children can have following possibilities of sexes

GGG - 3 girls
GGB - 2 girls one boy
GBG
GBB
BGG
BGB
BBG
BBB

Total possibilities of both sexes = 8

Number of choices where there are exactly 2 girls and 1 boy are:
GGB
GBG
BGG
=3.

Thus, probability = 3/8

Ans: "D"

This problem is similar to having exactly 2 heads in 3 tosses.

Alternate way;
The total number of possibilities = (Number of possible outcomes in each flip)^(Number of tosses) = 2^3=8

Likewise;
The total number possible sexes for 3 children = (Number of possible sex for each child)^(Number of children)
Number of possible sex for each child = 2 = (Boy or Girl)
Number of children = 3

Total possible sexes = 2^3 = 8

Possibilities to have exactly 2 Girls out of 3 child and 1 Boy out of remaining 1 Child:
=
C^{3}_{2}*C^{1}{1} = 3

Thus, probability = Favorable/total outcomes = 3/8.


Thanks. This explanation is much simpler than the one in the source. I've been using the F/T rule in most of my previous problems.

But the explanation in the source confused me.
_________________

petrifiedbutstanding

Re: Family with three children   [#permalink] 06 Feb 2011, 02:59
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