A die is tossed thrice. Find the probability of getting an odd number at least once.
Probability of getting an odd number in a single throw of a die $=\frac{3}{6}=\frac{1}{2}$
Similarly, probability of getting an even number $=\frac{3}{6}=\frac{1}{2}$
Probability of getting an even number three times $=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$
Therefore, probability of getting an odd number at least once
$=1-$ probability of getting an odd number in none of the throws
$=1 -$ probability of getting an even number thrice
$=1-\frac{1}{8}$
$=\frac{7}{8}$
If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random, then the probability that $2$ white and $1$ black ball will be drawn is
Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$
The probability that at least one of the events $A$ and $B$ occurs is $3/5$. If $A$ and $B$ occur simultaneously with probability $1/5$, then $P(A') + P(B')$ is
If $P\,(A) = \frac{1}{4},\,\,P\,(B) = \frac{5}{8}$ and $P\,(A \cup B) = \frac{3}{4},$ then $P\,(A \cap B) = $
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 $F )$