If X sim B(7, p) is a binomial variate and P( | vedclass

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

Question

(C) Given $X \sim B(n, p)$ with $n=7$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $P(X=3) = P(X=5)$, we have:$\bin

Vedclass · Accepted Answer

$\frac{5-\sqrt{15}}{2}$

Vedclass · Answer

(C) Given $X \sim B(n, p)$ with $n=7$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $P(X=3) = P(X=5)$, we have:$\binom{7}{3} p^3 (1-p)^{7-3} = \binom{7}{5} p^5 (1-p)^{7-5}$$\binom{7}{3} p^3 (1-p)^4 = \binom{7}{5} p^5 (1-p)^2$Since $\binom{7}{3} = \binom{7}{4} = 35$ and $\binom{7}{5} = \binom{7}{2} = 21$, we get:$35 p^3 (1-p)^4 = 21 p^5 (1-p)^2$Dividing both sides by $7 p^3 (1-p)^2$ (assuming $p 
eq 0, 1$):$5 (1-p)^2 = 3 p^2$$5 (1 - 2p + p^2) = 3 p^2$$5 - 10p + 5p^2 = 3p^2$$2p^2 - 10p + 5 = 0$Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{10 \pm \sqrt{100 - 4(2)(5)}}{2(2)} = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2}$Since $0 \le p \le 1$, we take $p = \frac{5 - \sqrt{15}}{2}$.

If $X \sim B(7, p)$ is a binomial variate and $P(X=3)=P(X=5)$ then $p=$

Similar Questions

The sum and product of the mean and variance of a binomial distribution are $82.5$ and $1350$ respectively. The number of trials in the binomial distribution is:

In a binomial distribution,the probability of success is $\frac{1}{4}$ and the standard deviation is $3$. Then,its mean is

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$,where $0 < p < 1.$ If $\frac{P(X = r)}{P(X = n - r)}$ is independent of $n$ and $r$,then

If the difference between the mean and variance of a binomial variate is $\frac{5}{9}$,then the probability of getting $2$ successes for an event when the experiment is conducted $5$ times,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module