If the probability function of a random varia | vedclass

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

Question

(C) For a probability distribution,the sum of all probabilities must be equal to $1$. Given $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$,we

Vedclass · Accepted Answer

$\log_3(\log_e 2)$

Vedclass · Answer

(C) For a probability distribution,the sum of all probabilities must be equal to $1$. Given $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$,we have: $\sum_{k=1}^{\infty} P(X=k) = \sum_{k=1}^{\infty} \frac{(3^c)^k}{k!} = 1$. Recall the Taylor series expansion for the exponential function: $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$. Therefore,$\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x - 1$. Substituting $x = 3^c$,we get: $e^{3^c} - 1 = 1$,which implies $e^{3^c} = 2$. Taking the natural logarithm on both sides: $3^c = \log_e 2$. Taking the logarithm base $3$ on both sides: $c = \log_3(\log_e 2)$. Thus,the correct option is $C$.

If the probability function of a random variable $X$ is given by $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$ (where $c$ is a constant),then $c =$

Similar Questions

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:

$A$ random variate $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X=3) = 2 P(X=1)$ and $P(X=2) = 0.3$,then $P(X=0)$ is equal to:

If a variable takes values $0, 1, 2, ..., n$ with frequencies proportional to the binomial coefficients $^nC_0, ^nC_1, ..., ^nC_n$,find the mean of the distribution.

The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module