91Ó°ÊÓ

Given,

The number of people who have been exposed to disease. $\left(n\right)=17$

The likelihood of developing the disease$\left(p\right)=40\%=0.40$

Used formula:

The probability formula for a binomial distribution is:

$P(X=x)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x(1-p{)}^{n-x}$

The likelihood that hospital capacity will be surpassed can be calculated as follows:

$$\begin{gathered} P(X>10)=\underset{11}{\overset{17}{∑}}\left(\begin{array}{c}17x\end{array}\right)(0.40{)}^x(1-0.40{)}^{17-x} \\ =P(X=11)+P(X=12)+….+P(X=17) \\ =\left[\begin{array}{l}\left(\left(\begin{array}{l}1711\end{array}\right)(0.40{)}^{11}(1-0.40{)}^{17-11}\right)+\left(\left(\begin{array}{c}1712\end{array}\right)(0.40{)}^{12}(1-0.40{)}^{17-12}\right)+…+\\ \left(\left(\begin{array}{c}1712\end{array}\right)(0.40{)}^{17}(1-0.40{)}^{17-17}\right)\end{array}\right] \\ =0.035 \end{gathered}$$

Therefore, the correct option is (b) that is$0.035$.

(a) The probability that the hospital's capacity will not be$0.011$.

(c) The probability that the hospital's capacity will not be $0.092$.

(d) The probability that the hospital's capacity will not be $0.965$.

(e) The probability that the hospital's capacity will not be$0.989.$