A batsman plays a match on a dry pitch with probability 0.6 and on a wet pitch with probability 0.4. He scores a century with probability 0.3 on a dry pitch and with probability 0.1 on a wet pitch. Consider the following statements:
(A) The probability that he scores a century is 0.22.
(B) If he scores a century, the probability that the pitch was dry is \(\displaystyle\frac{9}{11}\).
(C) The probability that he does not score a century is 0.78.
(D) If he does not score a century, the probability that the pitch was wet is 0.4.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (C) only
Let \(E_1\) and \(E_2\) be the events that the pitch is dry and wet, and C the event that he scores a century.
\(P(E_1)P(C|E_1) = 0.6\times 0.3 = 0.18\)
\(P(E_2)P(C|E_2) = 0.4\times 0.1 = 0.04\)
By the theorem of total probability:
\(P(C) = 0.18+0.04 = 0.22\)
\(P(C') = 1-0.22 = 0.78\)
By Bayes’ theorem:
\( \displaystyle P(E_1|C) = \frac{0.18}{0.22} = \frac{9}{11} \)
On a wet pitch he fails to score a century with probability 0.9:
\(P(E_2)P(C'|E_2) = 0.4\times 0.9 = 0.36\)
\( \displaystyle P(E_2|C') = \frac{0.36}{0.78} = \frac{6}{13} \)
So P(century) = 0.22, P(dry | century) = \(\frac{9}{11}\), P(no century) = 0.78 and P(wet | no century) = \(\frac{6}{13}\).
Checking the statements
(A) The probability that he scores a century is 0.22. — Correct. 0.18 + 0.04 = 0.22.
(B) If he scores a century, the probability that the pitch was dry is \(\frac{9}{11}\). — Correct. \(\frac{0.18}{0.22} = \frac{9}{11}\).
(C) The probability that he does not score a century is 0.78. — Correct. 1 − 0.22 = 0.78.
(D) If he does not score a century, the probability that the pitch was wet is 0.4. — Incorrect. It is \(\frac{0.36}{0.78} = \frac{6}{13}\); 0.4 is the chance of a wet pitch before the result is known.