If the function $P[X = x] = \begin{cases} \frac{K \cdot 2^x}{x!}, & x = 0, 1, 2, 3 \\ 0, & \text{otherwise} \end{cases}$ forms a probability mass function (p.m.f.),then the value of $K$ is:
- A$\frac{5}{19}$
- B$\frac{2}{19}$
- C$\frac{3}{19}$
- D$\frac{1}{19}$
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$A$ fair coin is tossed $2$ times. $A$ person receives $₹ X^{3}$ if he gets $X$ number of heads. His expected gain is $=$
The cumulative distribution function $F(x)$ of a discrete random variable $X$ is given by the following table:
$X = x$ $-1$ $0$ $1$ $2$ $F(X = x)$ $0.3$ $0.7$ $0.8$ $1$
Then $E(X^2) = $
| $X = x$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|
| $F(X = x)$ | $0.3$ | $0.7$ | $0.8$ | $1$ |
Then $E(X^2) = $
$A$ person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $X$ is observed, then the range of $X$ is
DifficultAP EAMCET 2004
View Solution$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$
Determine $P(X > 6)$.
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X)$ | $0$ | $k$ | $2k$ | $3k$ | $3k^2$ | $k^2$ | $2k^2$ | $7k^2+k$ |
Determine $P(X > 6)$.
Easy
View SolutionIf the probability distribution of a random variable $X$ is as follows, then the variance of $X$ is
$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\\hline P(X=x) & K & 2 K & 3 K^2 & K^2 \\\hline\end{array}$
$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\\hline P(X=x) & K & 2 K & 3 K^2 & K^2 \\\hline\end{array}$
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