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 $P \left( A ^{\prime} \cap B ^{\prime}\right)$.
Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$
In a hostel, $60 \%$ of the students read Hindi newspaper, $40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. A student is selected at random Find the probability that she reads neither Hindi nor English newspapers.
If $P(A) = 2/3$, $P(B) = 1/2$ and ${\rm{ }}P(A \cup B) = 5/6$ then events $A$ and $B$ are
Events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12}$ and $\mathrm{P}$ $($ not $ \mathrm{A}$ or not $\mathrm{B})=\frac{1}{4} .$ State whether $\mathrm{A}$ and $\mathrm{B}$ are independent?