Given:
The probability that a baby that is born is a boy is 0.52.
The probability that a baby that is born is a girl is 0.48.
To find:
The probability that the family has 0, 1, or 2 girls.
Explanation:
Using the binomial distribution,
[tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]Here,
[tex]\begin{gathered} n=2 \\ P(Birth\text{ of girls\rparen=}p=0.48 \\ P(B\imaginaryI rth\text{ of boys\rparen=}1-p=0.52 \end{gathered}[/tex]The probability that the family gets 0 girl child is,
[tex]\begin{gathered} P(X=0)=^2C_0(0.48)^0(0.52)^2 \\ =0.2704 \end{gathered}[/tex]The probability that the family gets 1 girl child is,
[tex]\begin{gathered} P(X=1)=^2C_1(0.48)^1(0.52)^1 \\ =0.2496 \end{gathered}[/tex]The probability that the family gets 2 girl children is,
[tex]\begin{gathered} P(X=2)=^2C_2(0.48)^2(0.52)^0 \\ =0.2304 \end{gathered}[/tex]So, the probability that the family has 0, 1, or 2 girls is,
[tex]\begin{gathered} P(E)=0.2704+0.2496+0.2304 \\ =0.7504 \end{gathered}[/tex]a) The tree diagram is,
b) The binomial distribution table for p(X) is,
