Question

a counter has 4 defective and 3 non defective items.If a sample of 2 items are down one after another without replacement what is the probablity that,the sample will most one defective

Answer 1

The probability that both items are defective:

${p_1} = \frac{4}{7} \cdot \frac{3}{6} = \frac{2}{7}$

Probability that only one item is defective:

${p_2} = \frac{4}{7} \cdot \frac{3}{6} + \frac{3}{7} \cdot \frac{4}{6} = \frac{4}{7}$

Then the wanted probability is

$p(x \ge 1) = {p_1} + {p_2} = \frac{{2 + 4}}{7} = \frac{6}{7}$

Answer: $\frac{6}{7}$