Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that One of them is black and other is red.
Total number of balls $=18$
Number of red balls $=8$
Number of black balls $=10$
Probability of getting first ball as red $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
Probability of getting second ball as black $=\frac{10}{18}=\frac{5}{9}$
Therefore, probability of getting first ball as black and second ball as red $=\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$
Therefore, probability that one of them is black and other is red
$=$ Probability of getting first ball black and second as red $+$ Probability of getting first ball red and second ball black
$=\frac{20}{81}+\frac{20}{81}$
$=\frac{40}{81}$
Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $P(B)=p .$ Find $p$ if they are independent.
The probabilities of occurrence of two events are respectively $0.21$ and $0.49$. The probability that both occurs simultaneously is $0.16$. Then the probability that none of the two occurs is
If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$
If $A$ and $B$ are arbitrary events, then
Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $\mathrm{P}(\mathrm{B})=p .$ Find $p$ if they are mutually exclusive.