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 $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{2}{3},$ then
In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted $NSS$ but not $NCC$.
If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$ find $P(A \cap B)$ if $A$ and $B$ are independent events
Two events $A$ and $B$ will be independent, if
If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$, then $x = $