probability

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Reva: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Fri Feb 19, 2010 11:52 am; Thanked: 1 times

by Reva » Sat Apr 09, 2011 3:06 pm

1) a couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OA is A

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Source: Beat The GMAT — Problem Solving | Sat Apr 09, 2011 3:06 pm

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sat Apr 09, 2011 7:40 pm

Reva wrote:1) a couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OA is A

#Total number of outcomes (i.e. number of ways to have have 4 boys or girls) = 2^4 = 16
#Favorable outcomes = Number of ways of choosing either two boys or two girls out of 4 = 4C2 = 6

So, the required probability = 6/16 = 3/8, A

Thanks
Anshu

(Every mistake is a lesson learned )

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Reva: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Fri Feb 19, 2010 11:52 am; Thanked: 1 times

by Reva » Sat Apr 09, 2011 7:48 pm

anshumishra wrote:
Reva wrote:1) a couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OA is A
#Total number of outcomes (i.e. number of ways to have have 4 boys or girls) = 2^4 = 16
#Favorable outcomes = Number of ways of choosing either two boys or two girls out of 4 = 4C2 = 6

So, the required probability = 6/16 = 3/8, A[/quot

can you please explain how did you get that total number of outcomes are 2^4.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sat Apr 09, 2011 7:54 pm

Reva wrote:
anshumishra wrote:
Reva wrote:1) a couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OA is A
#Total number of outcomes (i.e. number of ways to have have 4 boys or girls) = 2^4 = 16
#Favorable outcomes = Number of ways of choosing either two boys or two girls out of 4 = 4C2 = 6

So, the required probability = 6/16 = 3/8, A[/quot
can you please explain how did you get that total number of outcomes are 2^4.

Sure.
Total number of outcomes = (Number of possible outcomes for 1st child -> either boy or girl ) * (Number of possible outcomes for 2nd child -> either boy or girl) * (Number of possible outcomes for 3rd child -> either boy or girl) * (Number of possible outcomes for 4th child -> either boy or girl) = 2*2*2*2 = 2^4 = 16

Thanks
Anshu

(Every mistake is a lesson learned )

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Reva: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Fri Feb 19, 2010 11:52 am; Thanked: 1 times

by Reva » Sun Apr 10, 2011 6:31 am

thank you .

anshumishra wrote:
Reva wrote:
anshumishra wrote:
Reva wrote:1) a couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OA is A
#Total number of outcomes (i.e. number of ways to have have 4 boys or girls) = 2^4 = 16
#Favorable outcomes = Number of ways of choosing either two boys or two girls out of 4 = 4C2 = 6

So, the required probability = 6/16 = 3/8, A[/quot
can you please explain how did you get that total number of outcomes are 2^4.
Sure.
Total number of outcomes = (Number of possible outcomes for 1st child -> either boy or girl ) * (Number of possible outcomes for 2nd child -> either boy or girl) * (Number of possible outcomes for 3rd child -> either boy or girl) * (Number of possible outcomes for 4th child -> either boy or girl) = 2*2*2*2 = 2^4 = 16

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Sun Apr 10, 2011 7:00 am

Reva wrote:1) a couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

a. 3/8
b. 1/4
c. 3/16
d. 1/8
e. 1/16

OA is A

hi this question can also be thought as an experiment where 4 coins are thrown in air, and we've been asked to find out the probability of having 2 heads and 2 tails.

total no. of outcomes=2^4;
2heads and 2 tails can occur in 4!/2!*2!=6;
therefore required probability is 6/2^4=3/8 hence A

O Excellence... my search for you is on... you can be far.. but not beyond my reach!

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force5: Legendary Member; Posts: 582; Joined: Tue Mar 08, 2011 12:48 amFollowed by: 6 members; Thanked: 61 times; GMAT Score:740

by force5 » Sun Apr 10, 2011 12:06 pm

well said 6 possible scenarios are
BGGB
GBBG
BBGG
GGBB
GBGB
BGBG

6/16

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