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

An experiment succeeds twice as often as it fails. Then the probability,that in the next $6$ trials there will be at least $4$ successes,is

During the winter months,in a certain village in Scotland,the probability of a day having severe fog is $0.6$. The probability that in a given week there will be exactly two days with severe fog is:

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. Then the probability distribution of the number of jacks is:

For which binomial distribution are the mean $= 6$ and variance $= 2$?

Let a random variable $X$ have a Binomial distribution with mean $8$ and variance $4$. If $P(X \leqslant 2) = \frac{k}{2^{16}}$,then $k$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module