Let $X$ denote the number of hours you study during a randomly selected school day. The probability that $X$ can take the values $x$ has the following form,where $k$ is some unknown constant.
$P(X=x) = \begin{cases} 0.1, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(5-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
What is the probability that you study at least two hours? Exactly two hours? At most two hours?
$P(X=x) = \begin{cases} 0.1, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(5-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
What is the probability that you study at least two hours? Exactly two hours? At most two hours?
- A$0.75, 0.3, 0.55$
- B$0.55, 0.3, 0.75$
- C$0.3, 0.55, 0.75$
- D$0.75, 0.55, 0.3$
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The following table shows the probability distribution of smart phones sold in a shop per day:
Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$ Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$
Then $E(x) = ?$
| Number of smart phones $(x)$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|---|
| Probability $(P(x))$ | $k$ | $0.3$ | $0.15$ | $0.15$ | $0.1$ | $2k$ |
Then $E(x) = ?$
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View SolutionIf the probability that the random variable $X$ takes the value $x$ is given by $P(X=x) = k(x+1)3^{-x}$,for $x = 0, 1, 2, 3, \ldots$,where $k$ is a constant,then $P(X \geq 3)$ is equal to
DifficultJEE MAIN 2025
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