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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$A$ player tosses two coins. He wins $Rs. 10$ if $2$ heads appear,$Rs. 5$ if one head appears,and $Rs. 2$ if no head appears. Find the variance of the winning amount.
For the probability distribution given below,find $\operatorname{Var}(X)$.
$X$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $P(X=x)$ $0.07$ $0.2$ $0.3$ $k$ $0.07$ $0.04$ $0.02$
| $X$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ |
|---|---|---|---|---|---|---|---|
| $P(X=x)$ | $0.07$ | $0.2$ | $0.3$ | $k$ | $0.07$ | $0.04$ | $0.02$ |
$A$ coin is tossed three times. Let $X$ denote the number of times a tail follows a head. If $\mu$ and $\sigma^2$ denote the mean and variance of $X$, then the value of $64(\mu+\sigma^2)$ is :
DifficultJEE MAIN 2025
View SolutionLet a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$ $(a)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$
| Outcome | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $(a)$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |
Easy
View Solution$A$ fair die is tossed twice in succession. If $X$ denotes the number of fours in $2$ tosses,then the probability distribution of $X$ is given by
DifficultMHT CET 2023
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