Question

Suppose $A$ and $B$ are two equally strong table tennis players. Which of the following two events is more probable
$a)$ $A$ beats $B$ in exactly $3$ games out of $4$
$b)$ $A$ beats $B$ in exactly $5$ games out of $8$

• A. $a$
• B. $b$
• C. $a$ & $b$
• D. neither$a$ nor $b$

Answer 1

The correct option is A $a$

Since $A$ and $B$ are equally strong players,

the probability of $A^{\prime }s$ winning a game of table tennis is $\frac{1}{2}$.
$\therefore p=\frac{1}{2}$ and $q=\frac{1}{2}$ $(\because p+q=1)$
This is a case of binomial distribution
a) Here, $n=4,r=3,p=\frac{1}{2}$ and $P(X=r){=}^4{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{4-r}$
$\therefore P(X=3){=}^4{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^1=4\times \frac{1}{16}=\frac{1}{4}$
$\therefore P(A$ wins $3$ games out of $4)=\frac{1}{4}$
b) Now, $n=8,p=\frac{1}{2},q=\frac{1}{2},r=5$
$\therefore P(X=5){=}^8{C}_5{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^3=\frac{7}{32}$
$\therefore P(A$ wins $5$ games out of $8)=\frac{7}{32}$
Since $\frac{1}{4}>\frac{7}{32}$ therefore event (a) is more probable.