Suppose X follows a binomial distribution wit | vedclass

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

Question

(A) The probability mass function of a binomial distribution is given by $P(X = k) = {}^nC_k p^k (1 - p)^{n-k}$.We are given the ratio $\frac{P(X = r)

Vedclass · Accepted Answer

$p = \frac{1}{2}$

Vedclass · Answer

(A) The probability mass function of a binomial distribution is given by $P(X = k) = {}^nC_k p^k (1 - p)^{n-k}$.We are given the ratio $\frac{P(X = r)}{P(X = n - r)}$.Substituting the formula,we get:$\frac{P(X = r)}{P(X = n - r)} = \frac{{}^nC_r p^r (1 - p)^{n-r}}{{}^nC_{n-r} p^{n-r} (1 - p)^r}$.Since ${}^nC_r = {}^nC_{n-r}$,the binomial coefficients cancel out:$\frac{P(X = r)}{P(X = n - r)} = \frac{p^r (1 - p)^{n-r}}{p^{n-r} (1 - p)^r} = \frac{p^r}{(1 - p)^r} \cdot \frac{(1 - p)^{n-r}}{p^{n-r}} = \left( \frac{p}{1 - p} ight)^r \cdot \left( \frac{1 - p}{p} ight)^{n-r} = \left( \frac{p}{1 - p} ight)^r \cdot \left( \frac{p}{1 - p} ight)^{-(n-r)} = \left( \frac{p}{1 - p} ight)^{r - n + r} = \left( \frac{p}{1 - p} ight)^{2r - n}$.Alternatively,$\frac{P(X = r)}{P(X = n - r)} = \left( \frac{1 - p}{p} ight)^{n - 2r}$.For this expression to be independent of $n$ and $r$,the base must be $1$ (since $1^{	ext{anything}} = 1$).Thus,$\frac{1 - p}{p} = 1 \implies 1 - p = p \implies 2p = 1 \implies p = \frac{1}{2}$.

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

Similar Questions

$A$ 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

Let $X$ have a binomial distribution $B(n, p)$ such that the sum and the product of the mean and variance of $X$ are $24$ and $128$ respectively. If $P(X > n - 3) = \frac{k}{2^n}$,then $k$ is equal to.

One out of $9$ ships is likely to sink when they are set on sail. When $6$ ships are set on sail,the probability that exactly $3$ of them will not arrive safely is

$A$ fair coin is tossed $100$ times. The probability of getting tails an odd number of times is

$A$ binomial random variable $X$ satisfies $9 \cdot P(X=4) = P(X=2)$ when $n=6$. Then $p$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module