Question

A sample of $4$ items is drawn at random without replacement from a lot of $10$ items containing $3$ defective. If $X$ denotes the number of defective items in the sample then $P\left(0<x<3\right)$ is equal to

• A. $\frac{3}{10}$
• B. $\frac{4}{5}$
• C. $\frac{1}{2}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{4}{5}$

Explanation for the correct option:

Step 1. Since, the items are chosen without replacement.

$P(X=x)=\frac{\left({}^{3}C_x+{}^{7}C_{4-x}\right)}{{}^{10}C_4}$

According to the question, $P(0<X<3)$

Step 2. Substituting $x=1$ and $x=2$:

$P(0<X<3)=\left[\frac{({}^{3}C_1.{}^{7}C_3)}{210}+\frac{({}^{3}C_2.{}^{7}C_2)}{210}\right]$

$\Rightarrow \frac{3\times 35+3\times 21}{210}$

$\Rightarrow \frac{(105+63)}{210}$

$\Rightarrow \frac{168}{210}=\frac{\mathbf{4}}{\mathbf{5}}$

Hence, Option ‘B’ is Correct.