(C) The Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$,where $\lambda = 6$.We need to find $P(1 \leq X \leq 5) = P(X=1)

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(C) The Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$,where $\lambda = 6$.We need to find $P(1 \leq X \leq 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$.$P(1 \leq X \leq 5) = e^{-6} \left[ \frac{6^1}{1!} + \frac{6^2}{2!} + \frac{6^3}{3!} + \frac{6^4}{4!} + \frac{6^5}{5!} \right]$$= e^{-6} \left[ 6 + \frac{36}{2} + \frac{216}{6} + \frac{1296}{24} + \frac{7776}{120} \right]$$= e^{-6} [6 + 18 + 36 + 54 + 64.8]$$= e^{-6} [178.8]$$= 6 \times e^{-6} \left( \frac{178.8}{6} \right) = 6 \times e^{-6} (29.8)$.

Class 12 Mathematics Probability Probability distribution

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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:

AP EAMCET 2023 All AP EAMCET

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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$A$ random variable $X$ has the following probability distribution:
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$A$ fair six-faced die is rolled $12$ times. The probability that each face turns up exactly twice is equal to:

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If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is:
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$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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:

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