The probability of India winning a test match | vedclass

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

Question

(C) Let $p$ be the probability of winning a match,$p = \frac{1}{2}$.Let $q$ be the probability of losing a match,$q = 1 - p = \frac{1}{2}$.For India's

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) Let $p$ be the probability of winning a match,$p = \frac{1}{2}$.Let $q$ be the probability of losing a match,$q = 1 - p = \frac{1}{2}$.For India's second win to occur at the third test,India must win exactly one match in the first two tests and win the third test.The possible sequences for the first three matches are $(L, W, W)$ and $(W, L, W)$.The probability of the sequence $(L, W, W)$ is $q 	imes p 	imes p = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$.The probability of the sequence $(W, L, W)$ is $p 	imes q 	imes p = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$.The total probability is $\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$K$	$2K$	$2K$	$3K$	$K^2$	$2K^2$	$7K^2+K$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

The probability of India winning a test match against West Indies is $\frac{1}{2}$. Assuming independence from match to match,the probability that in a $5$ match series India's second win occurs at the third test,is

Similar Questions

$A$ random variable $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If $\operatorname{Var}(X) = E(X)$,then $n$ is:

$A$ person draws a card from a pack of $52$ cards and replaces it. After shuffling,he draws a card again. If he repeats this process,what is the probability that he draws a red card for the first time on the third draw?

The probability distribution of a random variable $X$ is given below:
$X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X=x)$ $0$ $K$ $2K$ $2K$ $3K$ $K^2$ $2K^2$ $7K^2+K$

Then,$P(0 < X < 5)$ is equal to:

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{x}{8}$ for $0 < x < 4$ and $f(x) = 0$ otherwise. Then $P(X \leq 2)$ is:

$A$ random variable $X$ has the following probability distribution:

$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Find the value of $P(0 < X < 6)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module