Question

The probability of India winning a test match against South Africa is $\dfrac{1}{2}$ assuming independence from a match played. The probability that in a $5$ match series India’s second win occurs at the third test is
A. $\dfrac{1}{8}$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{4}$
D. $\dfrac{2}{3}$

Answer 1

Hint: First, we need to analyze the given information carefully so that we are able to solve the given problem. Here, we are given some probabilities. Here in this question, we need to apply the formula of complementary probability and the probability of an event to obtain the required answer.
Formula to be used:
a) The formula when the events are opposite to each other is as follows.
$P\left( A \right) + P\left( {{A^{‘}}} \right) = 1$
b) The formula to calculate the probability of an event is as follows.
The probability of an event (say A),$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$

Complete step by step answer:
It is given that the probability that India wins is $\dfrac{1}{2}$ .
Let$P\left( A \right)$ be the probability that India wins then, $P\left( A \right) = \dfrac{1}{2}$
Now, we shall determine the probability that India loses.
Let $P\left( {A'} \right)$be the probability that India loses.
     We need to apply the formula complimentary of probability
$P\left( {A'} \right) = 1 - P\left( A \right)$
$ = 1 - \dfrac{1}{2}$
$ = \dfrac{{2 - 1}}{2}$
$ = \dfrac{1}{2}$
Hence $P\left( {A'} \right) = \dfrac{1}{2}$
     From the given information we can understand that India won the third test and it also won the first two matches and lost the other.
Therefore, the required probability$ = P\left( {A',A,A} \right) + P\left( {A,A',A} \right)$
$ = P\left( {A'} \right) \times P(A) \times P(A) + P(A) \times P(A') \times P(A)$
$ = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$
$ = {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^3}$
$ = \dfrac{1}{8} + \dfrac{1}{8}$
$ = \dfrac{2}{8}$
$ = \dfrac{1}{4}$
Hence, the desired probability is $\dfrac{1}{4}$

So, the correct answer is “Option C”.

Note: The probability of an event is nothing but the ratio of the number of favorable outcomes and the total number of outcomes. This is given by the formula$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$. We apply the complementary probability formula when the two events are opposite to each other. Hence, we got the required probability.