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

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

Question

(C) The probability mass function for a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$, where $q = 1-p$.Given $n=4$, the equation

Vedclass · Accepted Answer

$\frac{6}{7}$

Vedclass · Answer

(C) The probability mass function for a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$, where $q = 1-p$.Given $n=4$, the equation is $P(X=4) = 6 P(X=2)$.Substituting the formula: ${ }^4 C_4 p^4 q^0 = 6 \cdot { }^4 C_2 p^2 q^2$.Since ${ }^4 C_4 = 1$ and ${ }^4 C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$, we have:$1 \cdot p^4 = 6 \cdot 6 \cdot p^2 q^2$.$p^4 = 36 p^2 q^2$.Dividing both sides by $p^2$ (assuming $p 
eq 0$):$p^2 = 36 q^2$.Taking the square root of both sides:$p = 6q$ (since $p$ and $q$ are probabilities, they must be positive).Substitute $q = 1-p$:$p = 6(1-p)$.$p = 6 - 6p$.$7p = 6$.$p = \frac{6}{7}$.

For a binomial variate $X$, if $n=4$ and $P(X=4)=6 P(X=2)$, then the value of $p$ is

Similar Questions

Two cards are drawn successively with replacement from a fair deck of $52$ cards. Let $X$ denote the number of kings obtained when two cards are drawn. Then $E(X^2) = $

If $15$ coins are tossed,then the probability of getting $10$ heads is:

$A$ coin is tossed $10$ times. The probability of getting exactly six heads is

$A$ pair of fair dice is thrown $4$ times. If getting the same number on both dice is considered as a success,then the probability of exactly two successes is:

In a box containing $100$ apples,$10$ are defective. The probability that in a sample of $6$ apples,$3$ are defective is

Vedclass Test Series

Exam Paper Generator

Online Exam Module