The number of persons joining a cinema ticket counter in a minute follows a Poisson distribution with parameter $\lambda = 6$. The probability that at least one and at most five persons join the queue in a particular minute is:
- A$e^{-6 \times 6}(25.48)$
- B$e^{-6}\left(\frac{6}{2}+\frac{6^3}{3 !}+\frac{6^4}{4 !}\right)$
- C$6 \times e^{-6}(29.8)$
- D$e^{-6}\left(6+\frac{6^2}{2}+\frac{6^3}{3 !}+\frac{6^4}{4 !}\right)$
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At a telephone enquiry system,the number of phone calls regarding relevant enquiries follows a Poisson distribution with an average of $5$ phone calls during $10$-minute time intervals. The probability that there is at most one phone call during a $10$-minute time period is:
MediumAIEEE 2006
View SolutionGiven 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 $\alpha: \beta=$
Three fair coins,each numbered $1$ and $0$,are tossed simultaneously. The variance $\operatorname{Var}(X)$ of the probability distribution of the random variable $X$,where $X$ is the sum of the numbers on the uppermost faces,is:
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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