Given P(A \(\cup\) B)’ = 1/6, P(A \(\cap\) B) = 1/4 and P(A)’ = 1/4
\(\therefore\) P(A \(\cup\) B) = 1 – P(A \(\cup\) B)’
= 1 – \(1\over 6\) = \(5\over 6\)
and P(A) = 1 – P(A)’ = 1 – \(1\over 4\) = \(3\over 4\)
P(A \(\cup\) B) = P(A) + P(B) – P(A \(\cap\) B)
\(\implies\) \(5\over 6\) = \(3\over 4\) + P(B) – \(1\over 4\)
P(B) = \(1\over 3\)
\(\implies\) A and B are not equally likely,
Also, P(A \(\cap\) B) = P(A).P(B) = \(1\over 4\)
So, events are independent.
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
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is
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