$A$ service station manager sells gas at an average of ₹ $100$ per hour on a rainy day,₹ $150$ per hour on a dubious day,₹ $250$ per hour on a fair day,and ₹ $300$ per hour on a clear sky day. If weather bureau statistics show the probabilities of weather as follows,then his mathematical expectation is:
Weather Clear Fair Dubious Rainy Probability $0.50$ $0.30$ $0.15$ $0.05$
| Weather | Clear | Fair | Dubious | Rainy |
| Probability | $0.50$ | $0.30$ | $0.15$ | $0.05$ |
- A$257.5$
- B$252.5$
- C$250$
- D$247.5$
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Similar Questions
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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$ $(b)$ $1$ $0$ $0$ $0$ $0$ $0$
| Outcome | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
|---|---|---|---|---|---|---|
| $(b)$ | $1$ | $0$ | $0$ | $0$ | $0$ | $0$ |
Easy
View Solution$A$ random variable $X$ takes values $0, 1, 2, 3, \dots$ with probabilities $P(X=x) = k(x+1)\left(\frac{1}{2}\right)^x$. If $k$ is a constant,then $P(X=1) = $
In a box,there are $20$ cards,out of which $10$ are labelled as $A$ and the remaining $10$ are labelled as $B$. Cards are drawn at random,one after the other and with replacement,until a second $A$-card is obtained. The probability that the second $A$-card appears before the third $B$-card is
DifficultJEE MAIN 2020
View SolutionIf a variable takes values $0, 1, 2, ..., n$ with frequencies proportional to the binomial coefficients $^nC_0, ^nC_1, ..., ^nC_n$,find the mean of the distribution.
Difficult
View Solution$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $P(X = x)$ $0.15$ $0.23$ $0.10$ $0.12$ $0.20$ $0.08$ $0.07$ $0.05$
For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is
| $X = x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(X = x)$ | $0.15$ | $0.23$ | $0.10$ | $0.12$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is
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