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

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Manager
Joined: 19 Oct 2010
Posts: 179
Location: India
GMAT 1: 560 Q36 V31
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What is the probability for a family with three children to  [#permalink]

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06 Feb 2011, 01:20
1
3
00:00

Difficulty:

25% (medium)

Question Stats:

67% (01:03) correct 33% (01:13) wrong based on 149 sessions

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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

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petrifiedbutstanding

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Joined: 20 Dec 2010
Posts: 1820
Re: Family with three children  [#permalink]

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06 Feb 2011, 01:52
4
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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Manager
Joined: 19 Oct 2010
Posts: 179
Location: India
GMAT 1: 560 Q36 V31
GPA: 3
Re: Family with three children  [#permalink]

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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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Joined: 25 Apr 2014
Posts: 112
Re: What is the probability for a family with three children to  [#permalink]

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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?
Retired Moderator
Joined: 15 Jun 2012
Posts: 1017
Location: United States
What is the probability for a family with three children to  [#permalink]

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17 Sep 2014, 00:07
1
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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Joined: 25 Apr 2014
Posts: 112
Re: What is the probability for a family with three children to  [#permalink]

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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
Manager
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 201
Location: India
What is the probability for a family with three children to  [#permalink]

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09 Jul 2017, 20:58
Since the probability of a boy and a girl are equal the individual probability is 1/2 and since there are BGG , (1/2)^3 * 3!/2! that is we have to consider the arrangements = 3/8. Hope this method helps
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Re: What is the probability for a family with three children to  [#permalink]

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10 Jul 2017, 00:02
petrifiedbutstanding wrote:
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

This can be solved in 2 ways.. First by checking number of events or second, by probability

First
Family to have a boy and 2 girls.
Total numbers of events possible i s

GGG
GGB
GBG
BGG
GBB
BGB
BBG
BBB

So, probability = 3/8.

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Re: What is the probability for a family with three children to  [#permalink]

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12 Jul 2017, 16:03
petrifiedbutstanding wrote:
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

We need to determine the probability for a family with three children to have a boy and two girls.

P(G-G-B) = 1/2 x 1/2 x 1/2 = 1/8

We also see that there are 3 ways to arrange two girls and one boy:

G-G-B

G-B-G

B-G-G

Thus, the probability is 1/8 x 3 = 3/8.

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Joined: 05 Oct 2014
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Location: India
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GMAT 1: 580 Q41 V28
GPA: 3.8
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Re: What is the probability for a family with three children to  [#permalink]

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09 Oct 2017, 08:20
Such a great way to solve. Thnx

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.
Re: What is the probability for a family with three children to &nbs [#permalink] 09 Oct 2017, 08:20
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