Question

The probability of India winning a test match against West Indies is \[\dfrac{1}{2}\] . Assuming independence from match to match the probability that in a \[5\] match series India’s second win occurs at the third test is:
A) \[\dfrac{1}{8}\]
B) \[\dfrac{1}{4}\]
C) \[\dfrac{1}{2}\]
D) \[\dfrac{2}{3}\]

Answer 1

Hint: Here we will use the formula for finding the probability which states that the probability for occurring any event will be equals to Number of favorable outcomes divided by the total number of favorable/possible outcomes:
\[{\text{Probability}} = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}\]

Complete step-by-step solution:
Step 1: It is given in the question that the probability of India winning a test match is \[\dfrac{1}{2}\] . By relating this with the probability formula we can say that from \[5\] total matches, India will win \[2\] out of it.
Step 2: Since India’s second win of the match occurs at the third test, we will make cases out of it as shown below:
Case I: India will win the first and third match (W, L, W), where W represents won and L represents Loss. The probability of winning a match is \[\dfrac{1}{2}\] as given in the question and for not winning is also \[\dfrac{1}{2}\] because the sum of probability will be always \[1\].
\[ \Rightarrow {\text{Probability}} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}\]
By multiplying into the RHS side of the expression, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{1}{8}\]
Case II: India will win the second and third match (L, W, W).
\[ \Rightarrow {\text{Probability}} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}\]
By multiplying into the RHS side of the expression, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{1}{8}\]
Step 3: So, the final probability that in a \[5\] match series India’s second win occurs at the third test is:
\[ \Rightarrow {\text{Probability}} = \operatorname{P} \left( {{\text{case I}}} \right) + {\text{P}}\left( {{\text{case II}}} \right)\]
By substituting the values of probability of case I and II, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{1}{8} + \dfrac{1}{8}\]
By taking \[8\] common into the RHS side, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{2}{8}\]
By simplifying the term into the RHS side, we get:
\[ \Rightarrow {\text{Probability}} = \dfrac{1}{4}\]

\[\therefore \]Option B is correct.

Note: Students need to remember some basic points about the probability that the range of the probability will always lie between \[0 \leqslant {\text{P(A)}} \leqslant {\text{1}}\] .