If X is a Poisson variate such that alpha = P | vedclass

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

Question

(B) Given that $X$ is a Poisson variate with parameter $\lambda$,the probability mass function is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $

Vedclass · Accepted Answer

$\frac{\alpha}{3}$

Vedclass · Answer

(B) Given that $X$ is a Poisson variate with parameter $\lambda$,the probability mass function is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $\alpha = P(X=1) = P(X=2)$:$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$$\lambda = \frac{\lambda^2}{2} \Rightarrow \lambda = 2$ (since $\lambda > 0$).Now,calculate $\alpha$:$\alpha = P(X=1) = \frac{e^{-2} 	imes 2^1}{1!} = 2e^{-2}$.We need to find $P(X=4)$:$P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-2} 	imes 2^4}{24} = \frac{16 e^{-2}}{24} = \frac{2}{3} e^{-2}$.Since $\alpha = 2e^{-2}$,we have $e^{-2} = \frac{\alpha}{2}$.Substituting this into the expression for $P(X=4)$:$P(X=4) = \frac{2}{3} 	imes \frac{\alpha}{2} = \frac{\alpha}{3}$.

If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$,then $P(X=4)$ is equal to

Similar Questions

If the probability that an individual will suffer a reaction from an injection of a drug is $0.001$,then the probability that out of $2000$ individuals having that injection,more than $2$ individuals will suffer a reaction,is

In a game,$3$ coins are tossed. $A$ person is paid $₹150$ if he gets all heads or all tails,and he is supposed to pay $₹50$ if he gets one head or two heads. The amount he can expect to win or lose on an average per game in $₹$ is:

The cumulative distribution function (c.d.f.) $F(x)$ associated with the probability density function (p.d.f.) $f(x) = 3(1 - x^2)$ for $0 < x < 1$ and $f(x) = 0$ otherwise,is given by $F(x) = k(x - \frac{2x^3}{k})$. Find the value of $k$.

$A$ typist claims that he prepares a typed page with typo errors of $1$ per $10$ pages. In a typing assignment of $40$ pages,if the probability that the typo errors are at most $2$ is $p$,then $e^2 p=$

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that both balls are red.

Vedclass Test Series

Exam Paper Generator

Online Exam Module