Question

A box contains $10$ identical electronic components of which $4$ are defective. If $3$ components are selected at random from the box in succession, without replacing the units already drawn, what is the probability that two of the selected components are defective?

• A. $\frac{1}{5}$
• B. $\frac{5}{24}$
• C. $\frac{3}{10}$
• D. $\frac{1}{40}$

Answer 1

The correct option is C

$\frac{3}{10}$

Explanation for the correct answer:

Step 1: Express the given data:

Number of identical electronic components = $10$

Number of defective components = $4$

Number of non-defective components = $6$

Total number of ways of selecting three components

$$\begin{gathered} ={}^{10}C_3 \\ =120 \end{gathered}$$

Number of ways of selecting one non-defective component

$={}^{6}C_1$

Number of ways of selecting two defective components

$$\begin{gathered} ={}^{4}C_2 \\ =6 \end{gathered}$$

Step 2: Finding the probability of an event:

The probability that two of the selected components are defective

$$\begin{gathered} =\frac{{}^{6}C_1\times {}^{4}C_2}{{}^{10}C_3} \\ =\frac{6\times 6}{120} \\ =\frac{3}{10} \end{gathered}$$

Hence, option (C) is the correct answer.