Let A and B be two events such that [P( A B ) = 1 6,,P( A B ) = 1 4] and [P( A ) = 1 4,] w…

Question

Let $A$ and $B$ be two events such that $$P\left( {\overline {A \cup B} } \right) = {1 \over 6},\,P\left( { {A \cap B} } \right) = {1 \over 4}$$ and $$P\left( {\overline A } \right) = {1 \over 4},$$ where $$\overline A $$ stands for the complement of the event $A$. Then the events $A$ and $B$ are :

Accepted Answer

$$P(\overline {A \cup B} ) = {1 \over 6}$$

or, $$1 - P(A \cup B) = {1 \over 6}$$

$$\therefore$$ $$P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};$$ $$P(A \cap B) = {1 \over 4}$$; and $$P(\overline A ) = {1 \over 4};$$

$$\therefore$$ $$P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}$$

We know, $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

or, $${5 \over 6} = {3 \over 4} + P(B) - {1 \over 4}$$

or, $$P(B) = {5 \over 6} - {1 \over 2} = {1 \over 3}$$

Now, $$P(A)\,.\,P(B) = {3 \over 4}.\,{1 \over 3} = {1 \over 4} = P(A \cap B)$$

i.e., events A and B are mutually independent.

Since the probability of A and B are different, so they are not equally likely events.

Therefore, (A) is the correct option.

Solutions · 1