Question

The probability that an event will fail to happen is $0.05$ . The probability that the event will take place on $4$ consecutive occasions is
A. $0.00000625$
B. $0.18543125$
C. $0.00001875$
D. $0.81450625$

Answer 1

Hint:In this case, when the problem is based on probability, we must know some basic fundamentals of probability such as $P(E)$ is the probability of occurring an event and $P(not\,E)$ is the probability of not occurring an event. Here, we will first find out the probability that an event will happen and then use the formula ${\left\{ {P\left( E \right)} \right\}^n}$ to get the required probability.

Formula used:
The basic formula used for evaluating probability here is:
$Probability{\text{ }}of{\text{ }}getting{\text{ }}an{\text{ }}outcome = \dfrac{{Number{\text{ }}of{\text{ }}getting{\text{ }}favorable{\text{ }}outcomes}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}outcomes}}$
and, $Probability{\text{ }}\left( {of{\text{ }}getting{\text{ }}an{\text{ }}event} \right) = 1 - Probability{\text{ }}\left( {of{\text{ }}not\,getting{\text{ }}an{\text{ }}event} \right)$
i.e., $P(E)\, = 1 - P\left( {\overline E } \right)$
and the formula for calculating the probability that the event will take place in $n$ consecutive occasions is ${\left\{ {P\left( E \right)} \right\}^n}$.

Complete Step by step solution:
It is given that the Probability of event will fail to happen is $0.05$.
Let us consider the probability that event will happen is $P(E)$ and the probability that event will not happen is $P(not\,E)\,or\,P\left( {\overline E } \right)$.
Therefore, $P{\text{ }}\left( {Event{\text{ }}will{\text{ }}fail} \right){\text{ }} = P\left( {\overline E } \right) = 0.05$
Now, according to given formula
$P{\text{ }}\left( {Event{\text{ }}will{\text{ }}happen} \right){\text{ }} = 1 - P\left( {Event{\text{ }}will{\text{ }}fail} \right)$
or, $P(E)\, = 1 - P\left( {\overline E } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$
On substituting the value of $P\left( {\overline E } \right)$ in equation $(1)$, we get
$P{\text{ }}\left( {Event{\text{ }}will{\text{ }}happen} \right) = P(E)\, = 1 - P\left( {\overline E } \right)\,$
$P{\text{ }}\left( {Event{\text{ }}will{\text{ }}happen} \right) = 1 - 0.05 = 0.95$
But here we need to find out the probability that the event will take place in $4$ consecutive occasions for that we will use this formula: -
$Probability(that{\text{ }}the{\text{ }}event{\text{ }}will{\text{ }}take{\text{ }}place{\text{ }}in\,'n'\,\;consecutive{\text{ }}occasions) = {\left\{ {P\left( E \right)} \right\}^n}$
Thus, the required probability that the event will take place on $4$ consecutive occasions is ${\left\{ {P\left( {Event{\text{ will happen }}} \right)} \right\}^n} = {\left\{ {P\left( E \right)} \right\}^n}$
On substituting $n = 4$, we get
$Probability(that{\text{ }}the{\text{ }}event{\text{ }}will{\text{ }}take{\text{ }}place{\text{ }}in\,'4'\,\;consecutive{\text{ }}occasions) = {\left( {0.95} \right)^4}$
$ = 0.81450625$
Hence, the correct option is (D) $0.81450625$ .

Note: In this question, the probability at which event will fail is given. To determine the required probability, first we need to evaluate probability that event will happen by using the probability that event will fail and then use the formula ${\left\{ {P\left( E \right)} \right\}^n}$ to determine the probability in $n$ consecutive occasions.