The probability of India winning a test match against west Indies is $$\frac{1}{2}.$$ Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at third test is
A.
$$\frac{1}{8}$$
B.
$$\frac{1}{4}$$
C.
$$\frac{1}{2}$$
D.
$$\frac{2}{3}$$
Answer :
$$\frac{1}{4}$$
Solution :
Given that $$P$$ (India wins) $$= p =$$ $$\frac{1}{2}$$
∴ $$P$$ (India loses) $$= p’ =$$ $$\frac{1}{2}$$
Out of 5 matches india’s second win occurs at third test
⇒ India wins third test and simultaneously it has won one match from first two and lost the other.
∴ Required prob. = $$P (LWW) + P (WLW)$$
$$ = {\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^3} = \frac{1}{4}$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$