Let X be a binomial variate with parameters n | vedclass

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

Question

(B) Given that $X$ is a binomial variate with $n=6$ and probability of success $p$. Let $q = 1-p$ be the probability of failure.The probability mass f

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(B) Given that $X$ is a binomial variate with $n=6$ and probability of success $p$. Let $q = 1-p$ be the probability of failure.The probability mass function is given by $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$.We are given the condition $4 P(X=4) = P(X=2)$.Substituting the formula:$4 \cdot {}^{6}C_{4} p^{4} q^{6-4} = {}^{6}C_{2} p^{2} q^{6-2}$$4 \cdot {}^{6}C_{4} p^{4} q^{2} = {}^{6}C_{2} p^{2} q^{4}$Since ${}^{6}C_{4} = {}^{6}C_{2} = 15$, we can cancel them:$4 p^{4} q^{2} = p^{2} q^{4}$Dividing both sides by $p^{2} q^{2}$ (assuming $p, q 
eq 0$):$4 p^{2} = q^{2}$Taking the square root of both sides:$2p = q$Since $q = 1-p$, we have:$2p = 1-p$$3p = 1$$p = 1/3$

Let $X$ be a binomial variate with parameters $n=6$ and $p$. If $4 P(X=4)=P(X=2)$, then $p$ equals

Similar Questions

$A$ die is tossed twice. Getting a number greater than $4$ is considered a success. Then the variance of the probability distribution of the number of successes is

If a fair coin is tossed $8$ times, then the probability of getting at most $2$ heads is...

$A$ manufacturer of locks knows that $2 \%$ of his product is defective. If he sells the locks in boxes each with $100$ locks and guarantees that not more than $2$ locks will be defective in a box,then the probability that a box will fail to meet the guaranteed quality is

$A$ die is thrown twice. If getting a number greater than $4$ on the die is considered a success,then the variance of the probability distribution of the number of successes is

If $\mu$ and $\sigma^{2}$ are the mean and variance of a random variable $X$ whose probability mass function is given by $P(X=x) = \binom{6}{x} \left(\frac{1}{3}\right)^{x} \left(\frac{2}{3}\right)^{6-x}$ for $x = 0, 1, 2, \ldots, 6$,then the value of $2\mu + 12\sigma^{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module