Question

There are 25 calculators in a box. Two of them are defective. suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be include in the inspection?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{3}$

$\mathrm{Box}\begin{array}{c}2\mathrm{defective}\\ 23\mathrm{non}-\mathrm{defective}\end{array}$
Number of ways to select one defective and 4 non defective calculators.

$={}^2{C}_1{.}^{23}{C}_4$

$\therefore Requiredprobability$

$=\frac{{}^2{C}_1{.}^{23}{C}_4}{{}^{25}{C}_5}=\frac{2\times 8855}{53130}=\frac{1}{3}$