Each of the persons A and B independently tos | vedclass

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

Question

(C) Let $X$ be the number of heads obtained by person $A$ and $Y$ be the number of heads obtained by person $B$. Both $X$ and $Y$ follow a binomial di

Vedclass · Accepted Answer

$\frac{5}{16}$

Vedclass · Answer

(C) Let $X$ be the number of heads obtained by person $A$ and $Y$ be the number of heads obtained by person $B$. Both $X$ and $Y$ follow a binomial distribution $B(n=3, p=1/2)$.The probability of getting $k$ heads in $3$ tosses is given by $P(X=k) = \binom{3}{k} (1/2)^3 = \binom{3}{k} / 8$.We want to find $P(X=Y) = P(X=0, Y=0) + P(X=1, Y=1) + P(X=2, Y=2) + P(X=3, Y=3)$.Since $A$ and $B$ are independent,$P(X=k, Y=k) = P(X=k) 	imes P(Y=k) = [P(X=k)]^2$.$P(X=0) = \binom{3}{0}/8 = 1/8 \implies P(X=0, Y=0) = (1/8)^2 = 1/64$.$P(X=1) = \binom{3}{1}/8 = 3/8 \implies P(X=1, Y=1) = (3/8)^2 = 9/64$.$P(X=2) = \binom{3}{2}/8 = 3/8 \implies P(X=2, Y=2) = (3/8)^2 = 9/64$.$P(X=3) = \binom{3}{3}/8 = 1/8 \implies P(X=3, Y=3) = (1/8)^2 = 1/64$.Total probability $= 1/64 + 9/64 + 9/64 + 1/64 = 20/64 = 5/16$.

Each of the persons $A$ and $B$ independently tosses three fair coins. The probability that both of them get the same number of heads is:

Similar Questions

If three dice are thrown together,then the probability of getting $5$ on at least one of them is

$A$ contest consists of predicting the results (win,draw,or defeat) of $7$ football matches. $A$ sent his entry by predicting at random. The probability that his entry will contain exactly $4$ correct predictions is:

For a binomial distribution,the mean and variance are $6$ and $4$ respectively. Find the value of $n$.

In a binomial distribution,the mean is $15$ and the variance is $10$. Then the parameter $n$ is:

If $P$ and $Q$ each toss three coins,the probability that both get the same number of heads is

Vedclass Test Series

Exam Paper Generator

Online Exam Module