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 $NSS$.
Let $A$ be the event in which the selected student has opted for $NCC$ and $B$ be the event in which the selected student has opted for $NSS$.
Total number of students $=60$
Number of students who have opted for $NCC =30$
$\therefore $ $P(A)=\frac{30}{60}=\frac{1}{2}$
Number of students who have opted for $NSS =32$
$\therefore $ $P(B)=\frac{32}{60}=\frac{8}{15}$
Number of students who have opted for both $NCC$ and $NSS = 24$
$\therefore $ $P ( A$ and $B )=\frac{24}{60}=\frac{2}{5}$
$P ($ not $A$ and not $B)$
$= P(A ^{\prime}$ and $B ^{\prime})$
$= P \left( A^{\prime} \cap B ^{\prime}\right)$
$= P ( A \cup B )^{\prime}$ $[( A^{\prime} \cap B )=( A \cup B )^{\prime}$ by De Morgan's law $)]$
$=1- P ( A \cup B )$
$=1- P ( A$ or $B )$
$=1-\frac{19}{30}$
$=\frac{11}{30}$
Thus, the probability that the selected students has neither opted for $NCC$ nor $NSS$ is $\frac{11}{30}$
From the employees of a company, $5$ persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows :
S.No. | Name | Sex | Age in years |
$1.$ | Harish | $M$ | $30$ |
$2.$ | Rohan | $M$ | $33$ |
$3.$ | Sheetal | $F$ | $46$ |
$4.$ | Alis | $F$ | $28$ |
$5.$ | Salim | $M$ | $41$ |
A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over $35$ years?
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 opted for $NCC$ or $NSS$.
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.4$, $P ( A \cap B )=0.8$
For any two events $A$ and $B$ in a sample space