Answer:
a.S={hh,sh,hs,ss}
b.tex]\frac{1}{4}[/tex]
c.
Step-by-step explanation:
We are given that a sample space of three children
S={bbb,bbg,bgb,bgg,gbb,gbg,ggb,ggg}
a.We have to construct similar space for two children where h for healthy and s for sick.
Then the sample space of two children
S={hh,sh,hs,ss}
b.Number of cases favorable to two healthy children=1
Total number of cases=4
Number of cases for two healthy children=1
Probability =Number of favorable cases divided by total number of cases
Probability=
Hence, the probability of getting two healthy children=
c.We have to find the probability of getting exactly one healthy child and one sick child
Number of cases of one healthy child and one sick child={hs,sh}=2
Probability=
Hence, the probability of getting exactly one healthy child and one sick child=