91Ó°ÊÓ

A nursery advertises that it has10 elm trees for sale and 3 of the trees have already been infected with dutch elm disease.

The buying of trees will be done without replacement.

Out of 10 trees, 3 trees are already infected.Hence, the number of healthy trees is7

So, if a buyer purchases2 trees, the probability that both the 2 trees will be healthy will be given by,

$$\begin{gathered} \frac{\left(\begin{array}{c}7\\ 2\end{array}\right)\left(\begin{array}{c}3\\ 0\end{array}\right)}{\left(\begin{array}{c}10\\ 2\end{array}\right)}=\frac{{\displaystyle \frac{7!}{7!\left(7-2\right)!}}×{\displaystyle \frac{3!}{0!\left(3-0\right)!}}}{{\displaystyle \frac{10!}{10!\left(10-2\right)!}}} \\ \frac{\left(\begin{array}{c}7\\ 2\end{array}\right)\left(\begin{array}{c}3\\ 0\end{array}\right)}{\left(\begin{array}{c}10\\ 2\end{array}\right)}=\frac{21×1}{45} \\ =\frac{21}{45} \\ =0.40 \end{gathered}$$

Hence, whena buyer purchases 2 trees, the probability that both trees will be healthy is0.40

To calculate the probability that at leastone tree is infected, we have to subtractthe probability that both trees will be healthy when a buyer purchases 2 trees. That is,

$$\begin{gathered} 1-0.40 \\ =0.6 \end{gathered}$$

Hence, the probability that at least 1 of the trees is infected is 0.6.