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(A) The problem follows a Poisson distribution because the number of trials $n = 2000$ is large and the probability of success $p = 0.001$ is very sma

Vedclass · Accepted Answer

$\frac{4}{3 e^{2}}$

Vedclass · Answer

(A) The problem follows a Poisson distribution because the number of trials $n = 2000$ is large and the probability of success $p = 0.001$ is very small.For a Poisson distribution,the parameter $\lambda$ is given by $\lambda = n 	imes p$.$\lambda = 2000 	imes 0.001 = 2$.The probability mass function for a Poisson distribution is $P(X = x) = \frac{e^{-\lambda} \lambda^{x}}{x!}$.We need to find the probability for exactly $x = 3$ individuals.Substituting the values,we get $P(X = 3) = \frac{e^{-2} 	imes 2^{3}}{3!}$.$P(X = 3) = \frac{e^{-2} 	imes 8}{6} = \frac{4}{3 e^{2}}$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

If the probability that an individual will suffer a bad reaction from an injection is $0.001$,then the probability that out of $2000$ individuals,exactly $3$ individuals suffer a bad reaction is

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