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 none will fuse after$150$ days of use.

$$\left( 1 \right){\text{ 1 - }}{\left( {\dfrac{{19}}{{20}}} \right)^5} \\\ \left( 2 \right){\text{ }}{\left( {\dfrac{{19}}{{20}}} \right)^5} \\\ \left( 3 \right){\text{ }}{\left( {\dfrac{3}{4}} \right)^5} \\\ \left( 4 \right){\text{ 90}}{\left( {\dfrac{1}{4}} \right)^5} \\\$$

Answer 1

To solve the given problem first we have to use the given information that is
The probability that from given bulb produced by a factory will fuse after $150$ days of use $= 0.05$
Then to determine the probability that a bulb will not fuse after $150$ days of its use by using the formula:-
Probability of an event will not occur $= 1 -$probability of an event will occur
Now for $5$such bulbs use the probability function of binomial distribution which is
$P\left( X \right){ = ^n}{C_x}{q^{n - x}}{p^x}$ where$x = 1,2,3........n$
After putting in the values we get the answer.

Complete step by step answer:
It is given that probability that a bulb produced by a factory will fuse after $150$ days of its use $= 0.05$
Now to find the probability that a bulb will not fuse after $150$ days of its use, use the formula
Formula:-
Probability of an event that will not occur $= 1 -$ Probability of an event that will occur
The probability that a bulb will not fuse after $150$ days of its use $= 1 -$Probability that a bulb produced by a factory will fuse after $150$ days of its use
That is
The probability that a bulb will not fuse after $150$ days of its use $= 1 - 0.05$
$= 0.95$
Now to find the probability that no bulb will fuse after $150$ days of its use, use the probability function of binomial distribution which is
$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x}$where$x = 1,2,3........n$
Here it is asking for $5$ bulbs so $n = 5$
$X = x = 0$( asked in the question)
And $p = 0.05$(probability that a bulb will fuse after $150$ days of its use)
$q = 0.95$ (probability that a bulb will not fuse after $150$ days of its use)
Substituting all the values in the formula we get

$$P\left( {X = x} \right){ = ^n}{C_x}{q^{n - x}}{p^x} \\\ P\left( 0 \right){ = ^5}{C_0}{\left( {0.95} \right)^{5 - 0}}{\left( {0.05} \right)^0} \\\$$

On further solving we get

$$P\left( 0 \right){ = ^5}{C_0}{\left( {0.95} \right)^5}{\left( {0.05} \right)^0} \\\ { = ^5}{C_0}{\left( {\dfrac{{95}}{{100}}} \right)^5} \\\$$

Using the formula of Combination $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here $n = 5,r = 0$

$$P\left( 0 \right) = \dfrac{{5!}}{{\left( {0!} \right)\left( {5 - 0} \right)!}}{\left( {\dfrac{{95}}{{100}}} \right)^5} \\\ = \dfrac{{5!}}{{5!}}{\left( {\dfrac{{19}}{{20}}} \right)^5} \\\ = {\left( {\dfrac{{19}}{{20}}} \right)^5} \\\$$

Hence, $P\left( 0 \right) = {\left( {\dfrac{{19}}{{20}}} \right)^5}$
So, the probability that out of $5$ such bulbs none will fuse after $150$ days of its use is ${\left( {\dfrac{{19}}{{20}}} \right)^5}$

So, the correct answer is “Option 2”.

Note:
We cannot directly find the probability that out of $5$ such bulbs none will fuse without using the binomial distribution. One thing we should take care of is that we do not multiply $5$ with the probability that the $1$ bulb will not fuse.