Problem Solving

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

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

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.

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

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

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

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

6/16