In class $XI$ of a school $40\%$ of the students study Mathematics and $30 \%$ study Biology. $10 \%$ of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Let $A$ be the event in which the selected student studies Mathematics and $B$ be the event in which the selected student studies Biology.
Accordingly, $P ( A )=40 \%=\frac{40}{100}=\frac{2}{5}$
$P(B)=30 \%=\frac{30}{100}=\frac{3}{10}$
$P ( A$ and $B )=10 \%=\frac{10}{100}=\frac{1}{10}$
We know that $P ( A$ and $B )= P ( A )+ P ( B )- P ( A $ and $B )$
$\therefore P(A $ or $ B)=\frac{2}{5}+\frac{3}{10}+\frac{1}{10}=\frac{6}{10}=0.6$
Thus, the probability that the selected student will be studying Mathematics or Biology is $0.6$.
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 neither $NCC$ nor $NSS$.
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 $A$ 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
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
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $A ).$