Question

178 Dutch elm disease. A nursery advertises that it has 10 elm treesfor sale. Unknown to the nursery, 3 of the trees have already been infected with Dutch elm disease and will die withina year.

a. If a buyer purchases 2 trees, what is the probability that bothtrees will be healthy?

b. Refer to part a. What is the probability that at least 1 of thetrees is infected?

Short answer

a. When a buyer purchases 2 trees, the probability that both trees will be healthy is

0.40

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

Step-by-step solution

Given information

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

Calculating the probability that both trees will be healthy

a.

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,

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

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

Calculating the probability that at least 1 tree is infected.

b.

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.