The odds in favour of A winning a game of tab | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The probability of $A$ winning a single game is $p = \frac{1}{1+2} = \frac{1}{3}$.The probability of $A$ losing a single game is $q = 1 - p = 1 -

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) The probability of $A$ winning a single game is $p = \frac{1}{1+2} = \frac{1}{3}$.The probability of $A$ losing a single game is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.Let $X$ be the number of games won by $A$ in $n=3$ games. $X$ follows a binomial distribution $B(n, p) = B(3, 1/3)$.The probability of $A$ winning at least two games is $P(X \ge 2) = P(X=2) + P(X=3)$.$P(X=2) = \binom{3}{2} p^2 q^1 = 3 	imes (\frac{1}{3})^2 	imes (\frac{2}{3}) = 3 	imes \frac{1}{9} 	imes \frac{2}{3} = \frac{6}{27}$.$P(X=3) = \binom{3}{3} p^3 q^0 = 1 	imes (\frac{1}{3})^3 	imes 1 = \frac{1}{27}$.$P(X \ge 2) = \frac{6}{27} + \frac{1}{27} = \frac{7}{27}$.

The odds in favour of $A$ winning a game of table tennis against $B$ are $1:2$. If $3$ games are to be played, then the probability of $A$ winning at least two games out of the three is..... (in $/27$)

Similar Questions

For a Binomial distribution,$n=6$,if $9 P(X=4)=P(X=2)$,then $q=$

In a binomial distribution,the mean is $18$ and the variance is $12$,then $p = . . . . . .$

When employees with smart phones are randomly selected,$50 \%$ use them for office purpose. The probability that out of $10$ users exactly $2$ of them use for office purpose is

Let a random variable $X$ have a Binomial distribution with mean $8$ and variance $4$. If $P(X \leqslant 2) = \frac{k}{2^{16}}$,then $k$ is equal to:

For $X \sim B(n, p)$,if $p=0.6$ and $E(X)=6$,then $\operatorname{Var}(X)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module