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

$=\frac{20}{81}+\frac{20}{81}$

$=\frac{40}{81}$

$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .

Let $A$ and $B$ be independent events such that $\mathrm{P}(\mathrm{A})=\mathrm{p}, \mathrm{P}(\mathrm{B})=2 \mathrm{p} .$ The largest value of $\mathrm{p}$, for which $\mathrm{P}$ (exactly one of $\mathrm{A}, \mathrm{B}$ occurs $)=\frac{5}{9}$, is :

- [JEE MAIN 2021]

Let $A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur together is $1/6$ and the probability that neither of them occurs is $1/3$. The probability of occurrence of $A$ is

The probability that $A$ speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact

- [IIT 1975]

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{{12}}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then

- [IIT 1993]