Question

The probability of A's a Tenis game against $B$ is $\frac{1}{3}$. Find the probability that A wins at least $1$ of the total $3$ set of games.

• A. $\frac{1}{19}$
• B. $\frac{19}{27}$
• C. $\frac{1q}{19}$
• D. $\frac{7}{19}$

Answer 1

Answer: (B)

$\frac{19}{27}$

Given: Probability of $A$ winning the tennis game against $B$ is $\frac{1}{3}$ and a total of $3$ games is played.

To find: the probability that $A$ wins at least $1$ of the total $3$ set of games

Probability of $A$ winning at least $1$ game $=$ P (A winning exactly $1$ game) $+$ P(A winning exactly $2$ games) $+$ P(A winning exactly $3$ games)

This is Bernoulli's' theorem, hence

P(Success in each trial) $={}^n{C}_k{p}^k(1-p{)}^{n-k}$, in this case $n=3,$ $p=\frac{1}{3}$ and $k=1,2,3$

$={}^3{C}_1{\left(\frac{1}{3}\right)}^1{(1-\frac{1}{3})}^{3-1}+{}^3{C}_2{\left(\frac{1}{3}\right)}^2{(1-\frac{1}{3})}^{3-2}+{}^3{C}_3{\left(\frac{1}{3}\right)}^3{(1-\frac{1}{3})}^{3-3}$

$=3\times \frac{1}{3}\times \frac{2^2}{3^2}+3\times \frac{2}{3^2}\times \frac{1}{3}+1\times \frac{1}{3^3}\times 1$

$=\frac{12}{27}+\frac{6}{27}+\frac{1}{27}=\frac{12+6+1}{27}=\frac{19}{27}$