Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that First ball is black and second is red.

Total number of balls $=18$

Number of red balls $=8$

Number of black balls $=10$

Probability of getting first ball black $=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after the first draw.

Probability of getting second ball as red $=\frac{8}{18}=\frac{4}{9}$

Therefore, probability of getting first ball as black and second ball as red $=\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$

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

- [AIEEE 2005]

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- [IIT 1993]

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