The formula for binomial probabilities is-
[tex]P(x)=\frac{n!}{(n-x)!\cdot x!}\cdot p^x\cdot q^{n-x}[/tex]According to the problem, n = 12, p = 0.42. Since the selection is 12 adults and the percentage is 42%.
Let's find q
[tex]q=1-p=1-0.42=0.58[/tex]Using a calculator, we have
[tex]P(3)\approx0.121[/tex]This means the probability of exactly three is 12.1%, approximately.
For (b) at least four, we have to find probability when x = 4, 5, 6, 7, 8, 9, 10, 11, 12. Using a calculator, we find each probability, then we sum.
[tex]P(\ge4)=P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12)[/tex][tex]\begin{gathered} P(4)=\frac{12!}{\left(12-4\right)!\cdot4!}\cdot\mleft(0.42\mright)^4\cdot\mleft(0.58\mright)^{12-4}\approx0.197 \\ P(5)=\frac{12!}{\left(12-5\right)!\cdot5!}\cdot\mleft(0.42\mright)^5\cdot\mleft(0.58\mright)^{12-5}\approx0.229 \\ P(6)=\frac{12!}{\left(12-6\right)!\cdot6!}\cdot\mleft(0.42\mright)^6\cdot\mleft(0.58\mright)^{12-6}\approx0.193 \\ P(7)=\frac{12!}{\left(12-7\right)!\cdot7!}\cdot\mleft(0.42\mright)^7\cdot\mleft(0.58\mright)^{12-7}\approx0.120 \\ P(8)=\frac{12!}{\left(12-8\right)!\cdot8!}\cdot\mleft(0.42\mright)^8\cdot\mleft(0.58\mright)^{12-8}\approx0.054 \\ P(9)=\frac{12!}{\left(12-9\right)!\cdot9!}\cdot\mleft(0.42\mright)^9\cdot\mleft(0.58\mright)^{12-9}\approx0.017 \\ P(10)=\frac{12!}{\left(12-10\right)!\cdot10!}\cdot\mleft(0.42\mright)^{10}\cdot\mleft(0.58\mright)^{12-10}\approx0.004 \\ P(11)=\frac{12!}{\left(12-11\right)!\cdot11!}\cdot\mleft(0.42\mright)^{11}\cdot\mleft(0.58\mright)^{12-11}\approx0.0005 \\ P(12)=\frac{12!}{\left(12-12\right)!\cdot12!}\cdot\mleft(0.42\mright)^{12}\cdot\mleft(0.58\mright)^{12-12}\approx0.00003 \end{gathered}[/tex]Now, we sum all.
[tex]P(\ge4)=0.197+0.229+0.193+0.120+0.054+0.017+0.004+0.0005+0.00003=0.815[/tex]At last, for (a) at most two, the probability would be the sum of probabilities when x = 1, 2.
[tex]\begin{gathered} P(1)=\frac{12!}{\left(12-1\right)!\cdot1!}\cdot\mleft(0.42\mright)^1\cdot\mleft(0.58\mright)^{12-1}\approx0.013 \\ P(2)=\frac{12!}{\left(12-2\right)!\cdot2!}\cdot\mleft(0.42\mright)^2\cdot\mleft(0.58\mright)^{12-2}\approx0.050 \end{gathered}[/tex]So, the sum would be
[tex]P(\leq2)=P(1)+P(2)=0.013+0.050=0.063[/tex]