Let 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 Probability $\omega_{1}$ $1/8$ $\omega_{2}$ $2/3$ $\omega_{3}$ $1/3$ $\omega_{4}$ $1/3$ $\omega_{5}$ $-1/4$ $\omega_{6}$ $-1/3$
| Outcome | Probability |
|---|---|
| $\omega_{1}$ | $1/8$ |
| $\omega_{2}$ | $2/3$ |
| $\omega_{3}$ | $1/3$ |
| $\omega_{4}$ | $1/3$ |
| $\omega_{5}$ | $-1/4$ |
| $\omega_{6}$ | $-1/3$ |
(NONE) For an assignment of probabilities to be valid,it must satisfy two conditions:
$1$. Each probability $P(\omega_{i})$ must be such that $0 \le P(\omega_{i}) \le 1$ for all $i$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum_{i=1}^{6} P(\omega_{i}) = 1$.
In the given assignment,we observe that $P(\omega_{5}) = -1/4$ and $P(\omega_{6}) = -1/3$.
Since these probabilities are negative,they violate the first condition $(0 \le P(\omega_{i}) \le 1)$.
Therefore,this assignment of probabilities is not valid.
$1$. Each probability $P(\omega_{i})$ must be such that $0 \le P(\omega_{i}) \le 1$ for all $i$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum_{i=1}^{6} P(\omega_{i}) = 1$.
In the given assignment,we observe that $P(\omega_{5}) = -1/4$ and $P(\omega_{6}) = -1/3$.
Since these probabilities are negative,they violate the first condition $(0 \le P(\omega_{i}) \le 1)$.
Therefore,this assignment of probabilities is not valid.
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$A$ random variable $X$ has the following probability distribution:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Find the value of $P(0 < X < 6)$.
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| $P(X=x)$ | $0$ | $k$ | $2k$ | $2k$ | $3k$ | $k^2$ | $2k^2$ | $7k^2+k$ |
Find the value of $P(0 < X < 6)$.
Given below is the distribution of a random variable $X$:
$X=x$ $1$ $2$ $3$ $4$ $P(X=x)$ $\lambda$ $2\lambda$ $3\lambda$ $4\lambda$
If $\alpha=P(X < 3)$ and $\beta=P(X>2)$,then $\alpha: \beta=$
| $X=x$ | $1$ | $2$ | $3$ | $4$ |
| $P(X=x)$ | $\lambda$ | $2\lambda$ | $3\lambda$ | $4\lambda$ |
If $\alpha=P(X < 3)$ and $\beta=P(X>2)$,then $\alpha: \beta=$
The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$
| $X = x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x)$ | $\frac{1}{10}$ | $k$ | $\frac{1}{5}$ | $2k$ | $\frac{3}{10}$ | $k$ |
If the following function is a probability density function of a random variable $X$, $f(x) = kx^2(1 - x)$ for $0 < x < 1$ and $f(x) = 0$ otherwise, then the value of $k$ is:
DifficultMHT CET 2026
View SolutionIn a Poisson distribution, if $\frac{P(X=5)}{P(X=2)}=\frac{1}{7500}$ and $\frac{P(X=5)}{P(X=3)}=\frac{1}{500}$, then the mean of the distribution is
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