Respuesta :
Step 1: Write out the formula for binomial distribution
[tex]P(x)=^nC_x\times p^x\times q^{n-x}[/tex]Where
[tex]\begin{gathered} p\Rightarrow\text{probability of success} \\ q\Rightarrow\text{probability of failure} \\ n\Rightarrow\text{ number of trails } \\ x\Rightarrow\text{ number of success required} \end{gathered}[/tex]Step 2: State out the parameters needed in the formula to find the probabilty
[tex]\begin{gathered} p=9\text{ \%=}\frac{9}{100}=0.09 \\ q=1-p=1-0.09=0.91 \\ n=12 \\ x\Rightarrow\le2\Rightarrow0,1,2 \end{gathered}[/tex]Step 3: The probability that at most 2 children live with their father only can be described as;
[tex]P(x\le2)=P(0)+P(1)+P(2)[/tex]Step 4: Find the probability of each number of successes required
[tex]\begin{gathered} P(0)=^{12}C_0\times(0.09)^0\times(0.91)^{12-0} \\ P(0)=1\times1\times0.322475487=0.322475487 \end{gathered}[/tex][tex]\begin{gathered} P(1)=^{12}C_1\times(0.09)^1\times(0.91)^{12-1} \\ =^{12}C_1\times(0.09)^1\times(0.91)^{11} \\ =12\times0.09\times0.354368667=0.38271816 \end{gathered}[/tex][tex]\begin{gathered} P(2)=^{12}C_2\times(0.09)^2\times(0.91)^{12-2} \\ =^{12}C_2\times(0.09)^2\times(0.91)^{10} \\ =66\times0.0081\times0.389416118=0.208181856 \end{gathered}[/tex]Step 5: Add all the number of successess required
[tex]\begin{gathered} P(x\le2)=0.322475487+0.38271816+0.208181856 \\ =0.913375503 \\ \approx0.913 \end{gathered}[/tex]Hence, the probability that at most 2 children live with their father only is 0.913
