Question

From a batch of $100$ items of which $20$ are defective, exactly two items are chosen, one at a time, without replacement. Calculate the probabilities that the second item chosen is defective.

• A. $\frac{2}{5}$
• B. $\frac{19}{100}$
• C. $\frac{1}{5}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{5}$

Let A={first item chosen is defective}, B ={second item chosen is defective}

Given:

Total number of items = 100.

Number of defective items = 20

To find:

the probabilities that second item chosen is defective.

$P\left(B\right)=P\left(B/A\right)P\left(A\right)+P\left(B/A^{\prime }\right)P\left(A^{\prime }\right)=\frac{20}{100}\times \frac{19}{99}+\frac{20}{99}\times \frac{80}{100}=\frac{198}{990}=\frac{1}{5}$