Question

A packet of 10 CD's contains 4 defected. The CD's are selected at random, one by one, examined and are not replaced. The probability that 7th CD is the last defective is

• A. $\frac{2}{21}$
• B. $\frac{4}{9}$
• C. $\frac{7}{27}$
• D. None of these

Answer 1

The correct option is A $\frac{2}{21}$

In a packet of $10$ CD's, $4$ are defective

P($3$ are defective in first $6$)$=\frac{{}^6{C}_3{}^4{C}_3}{{}^{10}{C}_6}$

$=\frac{80}{210}$

$=\frac{8}{21}$

P($7^{th}$ selected is the last defective/$3$ are defective in the first $6$)

$=\frac{1}{4}$

$\therefore$ P($7^{th}$ selected is the last defective)

P($7^{th}$ selected is the last defective n 3 are defective in first $6$)

$=$P($3$ are defective in first $6$). P($7^{th}$ selected is the last defective/$3$ are defective in the first $6$)

$=\frac{8}{21}\cdot \frac{1}{4}$

$=\frac{2}{21}$.