Question

Team $A$ has probability $\frac{2}{3}$ of winning whenever it plays. If $A$ plays $4$ games the probability that $A$ loses all the games is

• A. ${\left(\frac{2}{3}\right)}^4$
• B. ${\left(\frac{1}{3}\right)}^4$
• C. ${\left(\frac{3}{4}\right)}^4$
• D. ${\left(\frac{1}{4}\right)}^4$

Answer 1

Answer: (B)

${\left(\frac{1}{3}\right)}^4$

Probability of the team winning is $\frac{2}{3}$.
Hence the probability of the team losing is $\frac{1}{3}$
Now the probability of A playing 4 games and loosing all will be ${}^4{C}_4\left(\frac{1}{3}{)}^4\right(\frac{2}{3}{)}^0$

$=\frac{1}{3^4}$

$=\frac{1}{81}$