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# What is the probability for a family with three children to

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Senior Manager
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Kudos [?]: 42 [1] , given: 27

What is the probability for a family with three children to [#permalink]  06 Feb 2011, 01:20
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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
[Reveal] Spoiler: OA

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petrifiedbutstanding

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Re: Family with three children [#permalink]  06 Feb 2011, 01:52
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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.
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Senior Manager
Joined: 19 Oct 2010
Posts: 274
Location: India
GMAT 1: 560 Q36 V31
GPA: 3
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Kudos [?]: 42 [0], given: 27

Re: Family with three children [#permalink]  06 Feb 2011, 01:59
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.
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petrifiedbutstanding

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Re: Family with three children [#permalink]  23 Feb 2014, 17:11
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Kudos [?]: 30 [0], given: 1474

Re: What is the probability for a family with three children to [#permalink]  14 Sep 2014, 10:33
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.

Hi Fluke,
How does the combination of BGG and GBG vary? Why are we considering the order of kids here?
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Kudos [?]: 1701 [1] , given: 123

What is the probability for a family with three children to [#permalink]  17 Sep 2014, 00:07
1
KUDOS
maggie27 wrote:
Hi Fluke,
How does the combination of BGG and GBG vary? Why are we considering the order of kids here?

Hello maggie27

Not Fluke, but I may help.

A family has 3 children --> Let put the order: C1_C2_C3 (child 1_child 2_child3).
Each child could be boy or girl --> 2 ways for C1; 2 ways for C2; 2 ways for C3. We have total 2x2x2 = 8 ways. It means you considered the order already i.e. BGB # BBG. Thus, when you count the combination (1boy, 2 girls), you have to consider the order. So we have three orders that have 1 boy and 2 girls
BGG; GBG; GGB

=> probability = 3/8

Hope it's clear.
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Re: What is the probability for a family with three children to [#permalink]  17 Sep 2014, 07:04
pqhai wrote:
maggie27 wrote:
Hi Fluke,
How does the combination of BGG and GBG vary? Why are we considering the order of kids here?

Hello maggie27

Not Fluke, but I may help.

A family has 3 children --> Let put the order: C1_C2_C3 (child 1_child 2_child3).
Each child could be boy or girl --> 2 ways for C1; 2 ways for C2; 2 ways for C3. We have total 2x2x2 = 8 ways. It means you considered the order already i.e. BGB # BBG. Thus, when you count the combination (1boy, 2 girls), you have to consider the order. So we have three orders that have 1 boy and 2 girls
BGG; GBG; GGB

=> probability = 3/8

Hope it's clear.

Thanks pqhai
I get it now! Kudos to u
Re: What is the probability for a family with three children to   [#permalink] 17 Sep 2014, 07:04
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