Probability

Question

Of a group of children, 0.4 are boys and 0.6 are girls. Of the boys, 0.7 have brown eyes; of the girls, 0.4 have brown eyes. A child is selected at random from the group.

(a) Find the probability that the child is a girl.


(b) Find P(brown eyes | boy).


(c) Find the probability that the child is a boy with brown eyes.


(d) Find the probability that the child is a girl with brown eyes.


(e) Find the probability that the child has brown eyes.


(f) Find the probability that the child is a girl, given that the child has brown eyes. (Round your answer to three decimal places.)

Answer 1

Stephanie,

 

Let's make up some codes to describe a child.  Each child has two characteristics of interest in the question...sex and eye colour.  A two-string code, the first letter representing sex and the second eye-colour will do.  Let B=boy,

G=girl Y(for yes)=Brown-eyed and N=Not Brown-eyed.

 

The possible combinations are:  BY, BN, GY, GN and the probabilities of each must sum to 1.

 

We are told 0.4 are boys and 0.6 are girls. (0.4+0.6=1)

 

Of the boys, 0.7 have brown eyes...so P(BY)=0.4*0.7=0.28  and P(BN)=0.4-0.28=0.12

Of the girls, 0.4 have brown eyes...so P(GY)=0.6*0.4=0.24 and P(GN)=0.6-0.24=0.36

 

a/ 0.6 as stated in the question

b/ 0.7 as stated in the question

c/ P(BY)=0.28

d/ P(GY)=0.24

e/ P(GY)+P(BY)=0.52

f/  P(GY)/0.52=0.24/0.52=0.462 (3 decimal places)