Question

The probability of India winning a test match against the West Indies is 1/2 assuming independence from match-to-match. The probability that in a match series India's second win occurs at the third test, is
$
  {\text{A}}{\text{. }}\dfrac{1}{8} \\
  {\text{B}}{\text{. }}\dfrac{1}{4} \\
  {\text{C}}{\text{. }}\dfrac{1}{2} \\
  {\text{D}}{\text{. }}\dfrac{2}{3} \\
$

Answer 1

Hint: - AS given in the question India’s second win should come in the third test that means either India should win the first and third match or it should win the second and third match, so we have to find the probability of these two cases.

Complete step by step answer:
Now we proceed as,
Let ${A_1},{A_2}{\text{ and }}{A_3}$be the events of match winning in first , second and third matches respectively and whose probabilities are
$P\left( {{A_1}} \right) = P\left( {{A_2}} \right) = P\left( {{A_3}} \right) = \dfrac{1}{2}$
And hence we can say
$P\left( {{A_1}'} \right) = P\left( {{A_2}'} \right) = P\left( {{A_3}'} \right) = \dfrac{1}{2}$
So required probability is
$ = P\left( {{A_1}{A_2}'{A_3}} \right) + P\left( {{A_1}'{A_2}{A_3}} \right)$
Using the property of probability we can write
$ = P\left( {{A_1}} \right)P\left( {{A_2}'} \right)\left( {{A_3}} \right) + P\left( {{A_1}'} \right)p\left( {{A_2}} \right)p\left( {{A_3}} \right)$
$ = {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^3} = \dfrac{1}{8} + \frac{1}{8} = \dfrac{1}{4}$
Hence option B is the correct option.

Note: -Whenever we get this type of question the key concept of solving is we take to consider cases as written in hint and this type of question can be easily tackled by understanding which is my favorable event and how many are total events.