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.
Total number of balls $=18$
Number of red balls $=8$
Number of black balls $=10$
Probability of getting a red ball in the first draw $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
$\therefore$ Probability of getting a red ball in the second draw $=\frac{8}{18}=\frac{4}{9}$
Therefore, probability of getting both the balls red $=\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}$
Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $A ).$
For the three events $A, B$ and $C, P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs)= $P$ (exactly one of the events $C$ or $A$ occurs)= $p$ and $P$ (all the three events occur simultaneously) $ = {p^2},$ where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is
Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$
Let $A$ and $B$ be two events such that the probability that exactly one of them occurs is $\frac{2}{5}$ and the probability that $A$ or $B$ occurs is $\frac{1}{2}$ then the probability of both of them occur together is