For a binomial variate X with n=6,if P(X=4)=f | vedclass

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

Question

(D) Given,$P(X=4) = \frac{135}{2^{12}}$.Using the binomial probability formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:${}^6C_4 p^4 q^2 = \frac{135}{2^

Vedclass · Accepted Answer

$\frac{9}{8}$

Vedclass · Answer

(D) Given,$P(X=4) = \frac{135}{2^{12}}$.Using the binomial probability formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:${}^6C_4 p^4 q^2 = \frac{135}{2^{12}}$.Since ${}^6C_4 = 15$,we get $15 p^4 q^2 = \frac{135}{2^{12}}$.Dividing by $15$,$p^4 q^2 = \frac{9}{2^{12}} = \frac{3^2}{(2^6)^2} = \left(\frac{3}{64}ight)^2$.Taking the square root,$p^2 q = \frac{3}{64}$.Substituting $q = 1-p$,we have $p^2(1-p) = \frac{3}{64}$.By inspection,if $p = \frac{1}{4}$,then $p^2(1-p) = (\frac{1}{16})(\frac{3}{4}) = \frac{3}{64}$.Thus,$p = \frac{1}{4}$ and $q = \frac{3}{4}$.The variance of a binomial distribution is given by $npq$.Variance $= 6 	imes \frac{1}{4} 	imes \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

For a binomial variate $X$ with $n=6$,if $P(X=4)=\frac{135}{2^{12}}$,then its variance is

Similar Questions

Two dice are thrown successively $4$ times and getting a doublet is called a success. The probability of getting at least $1$ success is (in $/1296$)

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials,there will be at least $4$ successes.

If two coins are tossed $5$ times,what is the probability of getting $5$ heads and $5$ tails?

If $7$ different balls are distributed among $4$ different boxes,then the probability that the first box contains $3$ balls is

Let $X \sim B(6, 1/2)$. Then the maximum probability occurs at

Vedclass Test Series

Exam Paper Generator

Online Exam Module