Question

If the probability of X to fail in the examination is 0.3 and that for Y is 0.2, then the probability that either X or Y fail in the examination is

• A. 0.5
• B. 0.44
• C. 0.6
• D. None of these

Answer 1

Answer: (B)

0.44

Answer 2

Here $P ( X ) = 0.3$ $P ( Y ) = 0.2$

Now $P ( X \cup Y ) = P ( X ) + P ( Y ) - P ( X \cap Y )$

Since these are independent events, so

$P ( X \cap Y ) = P ( X ) \cdot P ( Y )$

Thus required probability$= 0.3 + 0.2 - 0.06 = 0.44$.