The c.d.f. $F(x)$ associated with the p.d.f. $f(x)$ is given by:
$f(x) = \begin{cases} 12x^2(1-x), & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$
$f(x) = \begin{cases} 12x^2(1-x), & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$
- A$F(x) = 4x^3 + 3x^4$
- B$F(x) = 4x^3 - 3x^4$
- C$F(x) = -4x^3 - 3x^4$
- D$F(x) = -4x^3 + 3x^4$
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In a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is
$A$ random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3$, then $P(X = 0)$ is
DifficultMHT CET 2026
View SolutionIf the probability mass function (p.m.f.) of a random variable $X$ is given by the table below,then find the value of $k$.
$X$ $1$ $2$ $3$ $4$ $5$ $P(X = x)$ $k$ $\frac{k}{3}$ $\frac{k}{4}$ $\frac{k}{2}$ $\frac{k}{2}$
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X = x)$ | $k$ | $\frac{k}{3}$ | $\frac{k}{4}$ | $\frac{k}{2}$ | $\frac{k}{2}$ |
$A$ random variable $X$ has the following probability density function (p.d.f.):
$f(x) = kx(1-x), 0 \leqslant x \leqslant 1$
If $P(X > a) = \frac{20}{27}$,then find the value of $a$.
$f(x) = kx(1-x), 0 \leqslant x \leqslant 1$
If $P(X > a) = \frac{20}{27}$,then find the value of $a$.
Suppose three coins are tossed simultaneously. If $X$ denotes the number of heads,then the probability distribution of $X$ is:
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