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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.
_________________

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.
_________________

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

Show Tags

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?

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

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17 Sep 2014, 00:07

1

This post received 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.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

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

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

Show Tags

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

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