Question

There are $4$ defective items in a lot of $15$ items. The items are selected one by one at random without replacement till the last defective item is drawn. The probability that the 10th item examined is the last defective item is

• A. $\frac{1}{65}$
• B. $\frac{2}{65}$
• C. $\frac{3}{65}$
• D. $\frac{4}{65}$

Answer 1

The correct option is D $\frac{4}{65}$

The ${10}^{th}$ item has to be the $4^{th}$ defective term.

The $9$ previous tails should have $3$ defective items.

so, we place $3$ defective items $,m9$ places $m\frac{9}{3}$ ways.

The remaining $6$ items are non defective.

$\therefore$ Probability of picjing $6$ non defective and $4$ defective items

$\Rightarrow \frac{11}{15}\times \frac{10}{14}\times \frac{9}{13}\times \frac{9}{12}\times \frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}$

$\Rightarrow \frac{2}{15\times 14\times 13}\Rightarrow \frac{1}{13\times 15\times 1}\Rightarrow \frac{1}{91\times 15}$

required probability $\Rightarrow {}^9{C}_3\times \frac{1}{91\times 15}$

$\Rightarrow \frac{9!}{6!\times 3!}\times \frac{1}{91\times 15}=\Rightarrow \frac{9\times 8\times 7}{6\times 91\times 15}$

$\Rightarrow \frac{4}{65}$

$\therefore$ Probabilitr that the ${10}^{th}$ examined item is the last defective is $\frac{4}{65}$