If a discrete random variable $X$ takes the values $1, 2, 3, 4$ such that $2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4)$, then $P(X = 4) = ...$ (in $61$)
- A$15$
- B$30$
- C$10$
- D$6$
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Similar Questions
$A$ random variable $X$ has the following probability distribution:
$X = x$ $0$ $1$ $2$ $P(X = x)$ $4k - 10k^2$ $5k - 1$ $3k^3$
Then $P(X < 2)$ is:
| $X = x$ | $0$ | $1$ | $2$ |
| $P(X = x)$ | $4k - 10k^2$ | $5k - 1$ | $3k^3$ |
Then $P(X < 2)$ is:
In a Poisson distribution with unit mean,calculate the value of $\sum_{x=0}^{\infty} |x-\bar{x}| P(X=x)$,where $\bar{x}$ is the mean of the distribution.
If the probability mass function (p.m.f.) is given by $P(X) = k \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$ and $k > 0$,and $P(X) = 0$ otherwise,then the value of $k$ is:
$A$ random variable $X$ takes the values $0, 1$ and $2$. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$, then the mean of the random variable $X$ is
Two numbers are selected at random from the first six positive integers. If $X$ denotes the larger of the two numbers,then $\operatorname{Var}(X) = $
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