Question

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

Answer 1

Hint: To find the probability of India’s second win occurring at the third test match, India must win exactly one of the first two matches. Thus India can either win the first match , along with not winning the second match or not win the first match but winning in the second match. The probability of an event not happening is equal to the one minus probability of that event happening.

Complete step-by-step answer:
Let us assume that \[P\left( {{M_1}} \right)\] represents the probability of India winning the first Test match against Australia, \[P\left( {{M_2}} \right)\] be the probability of India winning the second Test match against Australia and\[P\left( {{M_3}} \right)\] be the probability of India winning the third Test match against Australia.
Since it is given that the probability of India winning a test match against Australia is $\dfrac{1}{2}$, independent of the match number, we can say that \[P\left( {{M_1}} \right) = P\left( {{M_2}} \right) = P\left( {{M_3}} \right) = \dfrac{1}{2}\].
Let us assume that \[P\left( {{M_1}^\prime } \right)\] represents the probability of India not winning the first Test match against Australia, \[P\left( {{M_2}^\prime } \right)\] be the probability of India not winning the second Test match against Australia and\[P\left( {{M_3}^\prime } \right)\] be the probability of India not winning the third Test match against Australia.
Since it is given that the probability of India winning a test match against Australia is $\dfrac{1}{2}$, independent of the match number, we can say that \[P\left( {{M_1}^\prime } \right) = P\left( {{M_2}^\prime } \right) = P\left( {{M_3}^\prime } \right) = 1 - \dfrac{1}{2}\].
We have to find the probability of India scoring its second win in the series in the third match. Therefore, India must win exactly one of the first two matches. Thus India can either win the first match , along with not winning the second match or not win the first match but winning in the second match.
Mathematically
\[P\left( {{M_1} \cap {M_2}^\prime \cap {M_3}} \right){\text{ or }}P\left( {{M_1}^\prime \cap {M_2} \cap {M_3}} \right)\]
\[P\left( {{M_1} \cap {M_2}^\prime \cap {M_3}} \right) \cup P\left( {{M_1}^\prime \cap {M_2} \cap {M_3}} \right)\]
The required probability is therefore can be calculated by substituting the values \[P\left( {{M_1}} \right) = P\left( {{M_2}} \right) = P\left( {{M_3}} \right) = \dfrac{1}{2}\] and \[P\left( {{M_1}^\prime } \right) = P\left( {{M_2}^\prime } \right) = P\left( {{M_3}^\prime } \right) = \dfrac{1}{2}\].
$
  P\left( {{M_1} \cap {M_2}^\prime \cap {M_3}} \right) \cup P\left( {{M_1}^\prime \cap {M_2} \cap {M_3}} \right) \\
  P\left( {{M_1}} \right)P\left( {{M_2}^\prime } \right)P\left( {{M_3}} \right) + P\left( {{M_1}^\prime } \right)P\left( {{M_2}} \right)P\left( {{M_3}} \right) \\
  \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{2}} \right) \\
= \dfrac{1}{8} + \dfrac{1}{8} \\
= \dfrac{2}{8} \\
= \dfrac{1}{4} \\
$
Thus the probability of India scoring its second win in the third match is $\dfrac{1}{4}$.

Note: The probability \[P\left( {A \cap B} \right)\] is equivalent to $P\left( A \right)P\left( B \right)$, for two independent events \[A\]and \[B\]. The probability of an event cannot exceed 1. The probability of an event not happening is equal to the one minus probability of that event happening.