Question

There are 4 defective items in a lot consisting of 10 items. From this lot, we select 5 items at random, The probability that there will be 2 defective items among them is -

• A. $\frac{1}{2}$
• B. $\frac{2}{5}$
• C. $\frac{5}{21}$
• D. $\frac{10}{21}$

Answer 1

Answer: (D)

$\frac{10}{21}$

Assuming that the question is to find probability of getting exactly 2 defective systems.
Number of ways in which $5$ items can be selected
= 10C5 $=252$
Number of ways in which $2$ defective systems and $3$ non-defective systems can be selected
= 4C2×6C3 $=6\times 20=120$
Required probability
$=120/152=10/21$