Question

Three players A, B and C play a game. The probability that A, B and C will finish the game are respectively $\frac{1}{2} , \frac{1}{3}$ and $\frac{1}{4}$ . The probability that the game is finished is

• A. $\frac{1}{8}$
• B. 1
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

$\frac{3}{4}$

Answer 2

We have, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{4}$ $\therefore$ Required probability $=1-P(\bar{A}) P(\bar{B}) P(\bar{C})$ $=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}$ $=1-\frac{1}{4}=\frac{3}{4}$