Question

The probability that an individual suffers a bad reaction from an injection is 0.001. The probability that out of 2000 individuals exactly three will suffer bad reaction is:
A.$\dfrac{2}{{{e}^{2}}}$\[\]
B. $\dfrac{2}{5{{e}^{2}}}$\[\]
C. $\dfrac{8}{5{{e}^{2}}}$\[\]
D.$\dfrac{4}{3{{e}^{2}}}$\[\]

Answer 1

Hint: The trials follow Poisson’s distribution. First, find the average rate of value from the given data of probability and number of trials. Then put the values in the Poisson distribution formula for 3 trials.
We know that a Poisson experiment is an experiment, which separates the outcomes into two categories, for example, profit or loss, success or failure, head or tail. The specialized form of the binomial distribution for the Poison experiment is called Poisson’s distribution.

Complete step-by-step solution:
The Poisson distribution can only be defined for events which are independent ( the probability of occurrence of one event does not affect the probability of occurrence of other events) and mutually exclusive(events do not occur at the same time. ) and also the average rate of events per time period is constant.
The probability distribution function for the random variable $x$ in Poisson distribution is given by
\[P\left( X=x \right)=\dfrac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}....(1)\]
Where the value of $x$ represents the number of events within the time period, $\lambda $ is the average rate of events within the time period and $e$ is the base of the natural logarithm. We know that the time period is infinity for Poisson distribution.
It is given in the question the event is suffering from an individual of a bad reaction from the injection, then the other event is experiencing a good reaction. The events of having a good and bad reaction from the injection cannot happen at the same time and do not affect the reaction of other individuals. So we can use the Poisson’s distribution here.
We know from binomial distribution of probability that the average rate of value $\lambda $ is the product of the probability of success $p$ and the number of events $n$. So $\lambda =np$. We see in the question there 2000 number of events and the probability of suffering from a bad reaction is 0.001. So $\lambda =np=2000\times 0.0001=2$
 We are asked to find the probability for $x=3$ number of a bad reaction. We put these values in equation(1) and get the required value as,
\[P\left( X=3 \right)=\dfrac{{{e}^{-2}}{{2}^{3}}}{3!}=\dfrac{4}{3{{e}^{2}}}\]
So the correct option is D.\[\]

Note: The key here is to find the value of $\lambda $ from the given data in the question. We should be careful of confusion between normal, exponential, and Poisson distribution as all of them involve exponential function. If the questions for probability more than one 3 trials we can subtract from 1, as $1-P\left( X<3 \right)$.