Question

The probability that at most 5 defective fuses will be found in a box of 200 fuses, if experience shows that 20% of such fuses are defective, is
A.$\dfrac{{{e^{ - 40}}{{40}^5}}}{{5!}}$
B.$\sum\limits_{x = 0}^5 {\dfrac{{{e^{ - 40}}{{40}^x}}}{{x!}}} $
C.$\sum\limits_{x = 6}^\infty {\dfrac{{{e^{ - 40}}{{40}^x}}}{{x!}}} $
D.$1 - \sum\limits_{x = 6}^\infty {\dfrac{{{e^{ - 40}}{{40}^x}}}{{x!}}} $

Answer 1

Hint: We will first write the given percentage of defective fuse as fraction. Then, use Poisson theorem to calculate the probability of fuses when there are a total of 200 bulbs. Poisson theorem states that $P\left( {X = x} \right) = \dfrac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}$, where $x$ is the number of times of an event, $\lambda $ is the mean.

Complete step-by-step answer:
We are given that 20% fuses are defective.
We will write the probability of defective fuses as fraction.
That is we can write the probability of defective fuses as $p = \dfrac{{20}}{{100}}$
Whenever we are given the probability of an event occurring for a unit and we want to find the probability of the event happening a certain number of times, then we calculate it using Poisson Theorem.
Poisson theorem states that $P\left( {X = x} \right) = \dfrac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}$, where $x$ is the number of times of an event, $\lambda $ is the mean.
We can calculate $\lambda $ as $\lambda = np$, where $n$ is the total number of units and $p$ is the probability of an event.
Here, $n = 200$ and $p = \dfrac{{20}}{{100}}$
Then, mean for the given condition is $200 \times \dfrac{{20}}{{100}} = 40$
We have to find the probability when at most 5 defective fuses will be found in a box of 200 fuses.
Then, we have to take the sum when there is no defective fuse, 1 defective fuse, 2 defective fuses, 3 defective fuses, 4 defective fuses and 5 defective fuses.
Hence, the probability is given by $\sum\limits_{x = 0}^5 {\dfrac{{{e^{ - 40}}{{40}^x}}}{{x!}}} $
Thus, option B is correct.

Note: For using the Poisson distribution, the rate of occurrence should be constant. Many students make mistakes by taking only $x = 5$ in the formula of Poisson distribution, but we have to find the probability of at most 5 defective fuses. Hence, we will find summation of all the possible cases.