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 $=\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$
If $E$ and $F$ are events such that $P ( E )=\frac{1}{4}$, $P ( F )=\frac{1}{2}$ and $P(E$ and $F )=\frac{1}{8},$ find : $P ( E$ or $F )$
$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively
If $P(B) = \frac{3}{4}$, $P(A \cap B \cap \bar C) = \frac{1}{3}{\rm{ }}$ and $P(\bar A \cap B \cap \bar C) = \frac{1}{3},$ then $P(B \cap C)$ is
Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that both balls are red.
In a horse race the odds in favour of three horses are $1:2 , 1:3$ and $1:4$. The probability that one of the horse will win the race is