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)

$\frac{10}{21}$

Answer 2

Total number of items = 10
Number of defective items = 4
Number of non-defective items = 10 - 4 = 6

We select 5 items at random from the lot of 10 items.
Total number of ways to select 5 items from 10 = $\binom{10}{5}$.

We are interested in the probability that there will be exactly 2 defective items among the 5 selected items.
If there are exactly 2 defective items among the 5 selected, then the remaining $5 - 2 = 3$ items must be non-defective.

Number of ways to select 2 defective items from the 4 defective items = $\binom{4}{2}$.
Number of ways to select 3 non-defective items from the 6 non-defective items = $\binom{6}{3}$.

The number of ways to select exactly 2 defective items and exactly 3 non-defective items is the product of the number of ways to select each type of item:
Number of favorable outcomes = $\binom{4}{2} \times \binom{6}{3}$.

The probability of selecting exactly 2 defective items among the 5 selected items is the ratio of the number of favorable outcomes to the total number of outcomes:
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{\binom{4}{2} \times \binom{6}{3}}{\binom{10}{5}} = \frac{6 \times 20}{252} = \frac{120}{252} = \frac{10}{21}$.