Question

A coin is tossed $100$ times in which head is obtained $55$ times. One tossing a coin at random, find the probability of getting
(i) a head
(ii) a tail

Answer 1

Hint: In this problem we need to calculate the probabilities of the given two conditions based on the given data. In the problem we have given that the coin is tossed $100$. So the total number of events in the sample space will become $100$. They have also mentioned that when a coin is tossed $100$ times, heads are obtained $55$ times. So the number of favorable conditions for the head will become $55$. We know that the probability is the ratio of the number of favorable conditions to the total number of events in the sample space. From this we can calculate the probability of getting a head. Now the probability of getting a tail is the complementary event of getting a head. So the sum of the probabilities of the two events is equal to $1$ . Based on this statement we can frame a mathematical equation from which we will calculate the probability of getting a tail.

Complete step by step answer:
(i)
Given that, a coin is tossed $100$ times.
Hence the total number of events in the sample space is equal to $100$.
$n\left( S \right)=100$
Let us assume that $A$ be the event of getting a head while tossing a coin $100$ times.
In the problem we have given that when a coin is tossed $100$ times, the head has appeared $55$ times.
Hence the number of favorable cases for the event $A$ will be
$n\left( A \right)=55$
Now the probability of the event $A$ is given by
$\begin{align}
  & p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)} \\
 & \Rightarrow p\left( A \right)=\dfrac{55}{100} \\
 & \Rightarrow p\left( A \right)=\dfrac{11}{20} \\
\end{align}$
Hence the probability of getting a head when a coin is tossed $100$times is $\dfrac{11}{20}$ .

(ii)
Let $B$ be the event of getting a tail when a coin is tossed $100$ times.
The events $A$ and $B$ are mutually exclusive and complementary events in nature. So the sum of the probabilities of the both events is equal to one. So we will have
$p\left( A \right)+p\left( B \right)=1$
Substituting the value $p\left( A \right)=\dfrac{11}{20}$ in the above equation and simplifying the equation by using basic mathematical operations, then we will have
$\begin{align}
  & \dfrac{11}{20}+p\left( B \right)=1 \\
 & \Rightarrow p\left( B \right)=1-\dfrac{11}{20} \\
 & \Rightarrow p\left( B \right)=\dfrac{20-11}{20} \\
 & \Rightarrow p\left( B \right)=\dfrac{9}{20} \\
\end{align}$
Hence the probability of getting a tail when a coin is tossed $100$ times is $\dfrac{9}{20}$ .

Note: We can also calculate the number of favorable cases for getting a tail and from that value we can calculate the probability by calculating the ratio of favorable cases to the total number of events in the sample space. We will get favorable cases for getting a trial by subtracting favorable cases for getting a head from the total number of cases in the sample space.