Answer:
Fractional probability of having 3 boys and 0 girls in a family for first 3 births is [tex]\frac{1}{8}[/tex].
Step-by-step explanation:
Let A be the event of birth of a boy.
Let B be the event of birth of a girl.
Probability of birth of a boy, P(A) [tex]= \frac{1}{2}[/tex]
Probability of birth of a girl, P(B) [tex]= \frac{1}{2}[/tex]
We have to find the probability of birth of 3 boys i.e. event A occurring 3 times in a succession. It is independent events i.e. do not have any dependency on occurrence of any event.
So, Required probability can be calculated just by multiplying P(A) for 3 times.
i.e.
[tex]P(E) = P(A) \times P(A) \times P(A)\\\Rightarrow \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow \dfrac{1}{8}[/tex]
So, the probability of having 3 boys and 0 girls in the family is [tex]\frac{1}{8}[/tex].
