Question

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs, not more than one will fuse after 150 days of use.

Answer 1

Hint: Assume that after 150 days of use ‘X’ number of bulbs gets fused. Now, ‘X’ should not be greater than 1, according to the above question. Apply Bernoulli probability distribution formula given by: - \[P\left( X=a \right)={}^{n}{{C}_{a}}{{p}^{a}}{{q}^{n-a}}\] to calculate the probability of getting ‘a’ number of bulbs fused after 150 days. Here, ‘n’ is the number of bulbs available, ‘a’ is the number of bulbs getting fused, ‘p’ is the probability of getting the bulb fused and ‘q’ is the probability of not getting the bulb fused.

Complete step-by-step solution
We have been given that out of 5 bulbs, not more than one bulb should fuse. That means either 0 bulbs can fuse or 1 bulb can fuse after 150 days.
Now, applying Bernoulli probability distribution to find the probability of getting X = a bulbs fused, we have,
\[P\left( X=a \right)={}^{n}{{C}_{a}}{{p}^{a}}{{q}^{n-a}}\]
Here, p = probability of getting the bulb fused.
q = probability of not getting the bulb fused.
n = number of bulbs.
We have been given that, probability of getting the bulb fused will be (1 – 0.05) = 0.95. Hence,
p = 0.05, q = 0.95, n = 5.
Now, probability of the event that not more than one bulb will fuse.
\[\begin{align}
  & =P\left( X\le 1 \right) \\
 & =P\left( X=0 \right)+P\left( X=1 \right) \\
 & ={}^{5}{{C}_{0}}{{\left( 0.05 \right)}^{0}}{{\left( 0.95 \right)}^{5}}+{}^{5}{{C}_{1}}{{\left( 0.05 \right)}^{1}}{{\left( 0.95 \right)}^{5-1}} \\
 & ={{\left( 0.95 \right)}^{5}}+5\left( 0.05 \right){{\left( 0.95 \right)}^{4}} \\
 & ={{\left( 0.95 \right)}^{4}}\left[ 0.95+0.25 \right] \\
 & =1.2\times {{\left( 0.95 \right)}^{4}} \\
\end{align}\]

Note: One may note that it will be very difficult to solve the above problem without using the Bernoulli probability distribution rule. We have used p and q for denoting the probability of success and failure respectively because they are used as general notation in the formula. One can use different notations but do not get confused in the formula. ‘p’ must be raised to the power ‘a’ and ‘q’ to the power (n - a).