Question

What is the probability $P\left( E \right)$ of an event $E$ ?

Answer 1

Hint: We will solve this question using definition and the formula. $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$ $=$ $\dfrac{\text{no.of elements in E}}{\text{no.of elements in S}}$

Complete step by step answer:
For understanding the probability of an event, let us first understand some basic concepts.
Sample space: The set of all possible outcomes of a random experiment is called sample space for that experiment. It is usually denoted by S.
For example,
(i) When a coin is tossed, either a head or a tail will come up. If H denotes the occurrence of head and T denotes the occurrence of tail, then sample space \[S=\left\{ H,T \right\}\] .
(ii) When two coins are tossed, sample space \[S=\left\{ \left( H,H \right)\left( H,T \right)\left( T,H \right)\left( T,T \right) \right\}\] where $\left( H,H \right)$ denotes the occurrence of head on the first coin and occurrence of head on the second coin. Similarly, $\left( H,T \right)$ denotes the occurrence of head on the first coin and occurrence of tail on the second coin.
(iii) When a dice is thrown, any of the numbers $1,2,3,4,5,6$ will come up. Therefore, sample space\[S=\left\{ 1,2,3,4,5,6, \right\}\]. Here, $1$ denotes the occurrence of $1$, $2$ denotes the occurrence of $2$, and so on.
(iv) When two dice are thrown, sample space $S=\left\{ \begin{align}
  & \left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)\left( 1,4 \right)\left( 1,5 \right)\left( 1,6 \right)\left( 2,1 \right)\left( 2,2 \right)\left( 2,3 \right)\left( 2,4 \right)\left( 2,5 \right)\left( 2,6 \right) \\
 & \left( 3,1 \right)\left( 3,2 \right)\left( 3,3 \right)\left( 3,4 \right)\left( 3,5 \right)\left( 3,6 \right)\left( 4,1 \right)\left( 4,2 \right)\left( 4,3 \right)\left( 4,4 \right)\left( 4,5 \right)\left( 4,6 \right) \\
 & \left( 5,1 \right)\left( 5,2 \right)\left( 5,3 \right)\left( 5,4 \right)\left( 5,5 \right)\left( 5,6 \right)\left( 6,1 \right)\left( 6,2 \right)\left( 6,3 \right)\left( 6,4 \right)\left( 6,5 \right)\left( 6,6 \right) \\
\end{align} \right\}$, where $\left( 1,1 \right)$ denotes the occurrence of 1 on the first dice and occurrence of 1 on the second dice. Similarly, $\left( 1,2 \right)$ denotes the occurrence of 1 on the first dice and occurrence of 2 on the second dice.

We have learnt what a sample space is, let us learn about events.

Event: A subset of sample space S is called an event.
For example,
(i) When a coin is tossed, sample space \[S=\left\{ H,T \right\}\]. Let \[A=\left\{ H \right\}\], \[B=\left\{ T \right\}\] and \[C=\left\{ H,T \right\}\]. Here, \[A\] is the event of occurrence of a head, \[B\] is the event of occurrence of a tail, and \[C\] is the event of occurrence of a head or a tail.
(ii) When a die is thrown, sample space \[S=\left\{ 1,2,3,4,5,6, \right\}\]. Let \[A=\left( 1,3,5 \right)\]. Here \[A\] is the event of the occurrence of an odd number. Let \[B=\left( 5,6 \right)\]. Then, \[B\] is the event of occurrence of a number greater than 4. $C=\left( 2,3,5 \right)$. \[C\] is the event of the occurrence of a prime number.
(iii) When two coins are tossed, sample space \[S=\left\{ \left( H,H \right)\left( H,T \right)\left( T,H \right)\left( T,T \right) \right\}\]. Here $\left( H,T \right)$ denotes the event of the occurrence of a head on the first coin and tail on the second coin.

Now let us understand the probability of an event.
Let $S$ be the sample space, then probability of an event $E$ is denoted by $P\left( E \right)$.
\[\therefore P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}\] $=$ $\dfrac{\text{no.of elements in E}}{\text{no.of elements in S}}$
For example,
(i) When a coin is tossed, sample space\[S=\left\{ H,T \right\}\]. Let $E$ be the event of the occurrence of a head.
Then, $E=\left\{ H \right\}$
$\therefore P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{1}{2}$
(ii) When two coins are tossed, sample space\[S=\left\{ \left( H,H \right)\left( H,T \right)\left( T,H \right)\left( T,T \right) \right\}\]. Let $E$ be the event of occurrence of one head and one tail, then
$E=\left\{ \left( H,T \right)\left( T,H \right) \right\}$
$\therefore P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{2}{4}=\dfrac{1}{2}$

Note: The probability \[P\left( E \right)\] of occurrence of an event $E$ is a number lying between $0$ and $1$.i.e., $0\le P\left( E \right)\le 1$.