Solution
a.
[tex]\begin{gathered} n=15 \\ p=0.4 \\ \\ P(X=4)=C(15,4)\times(0.4)^4\times(0.6)^{11} \\ \\ P(X=4)=\frac{15!}{11!4!}\times(0.4)^4\times(0.6)^{11} \\ \\ P(X=4)=\frac{15\times14\times13\times12\times11!}{11!\times24}\times(0.4)^4\times(0.6)^{11} \\ \\ P(X=4)=1365\times(0.4)^4\times(0.6)^{11} \\ \\ P(X=4)=0.127 \end{gathered}[/tex]b.
[tex]\begin{gathered} n=12 \\ p=0.2 \\ \\ P(X=2)=C(12,2)\times0.2^2\times0.8^{10} \\ \\ P(X=2)=\frac{12!}{10!2!}\times0.2^2\times0.8^{10} \\ \\ P(X=2)=\frac{12\times11\times10!}{10!\times2}\times0.2^2\times0.8^{10} \\ \\ P(X=2)=66\times0.2^2\times0.8^{10} \\ \\ P(X=2)=0.283 \end{gathered}[/tex]C.
[tex]\begin{gathered} n=20 \\ p=0.05 \\ \\ P\left(x≤3\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right) \\ \\ P\left(x≤3\right)=0.984 \end{gathered}[/tex]