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

Hint: 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.