Question

The probability of A winning a game is $\frac{2}{3}$. In a series of 4 such games the probability that A will win more than half of the games is

• A. $32/81$
• B. $33/81$
• C. $16/27$
• D. $8/27$

Answer 1

Answer: (C)

$16/27$

Probability of A winning a game is$=\frac{2}{3}$
Probability of A losing a game is $=\frac{1}{3}$
Cases when A will win more than half of the games out of 4 is:
Case 1: $WWWL;WWLW;WLWW:LWWW$
Probability for case 1 $=4\times \left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{1}{3}\right)=\frac{32}{81}$
Case 2:$WWWW$
Probability for case 2 $=\left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)=\frac{16}{81}$
Probability that A will win more than half of the games is$=Case1+Case2=\frac{32}{81}+\frac{16}{81}=\frac{16}{27}$