Question

A coin is tossed two times, what is the probability of getting head at least once.

Answer 1

Hint: To solve this question first we need to list all the possible outcomes when a coin is tossed. Then we count the total outcomes and use the formula to calculate the probability. The following formula is used-
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes

Complete step by step solution:
As we know that the probability of an event is written as $P(A)$ where A is an event.
We have given that a coin is tossed two times. A coin has two faces i.e. Head and tail.
Let us assume that H represents Head and T represents Tail.
Now we list all the possible outcomes when we toss a coin two times.
$S=\left\{ H,H \right\},\left\{ H,T \right\},\left\{ T,H \right\},\left\{ T,T \right\}$.
We get the total number of possible outcomes $=4$
Now, we have to find the probability of getting head at least once, so the favourable outcomes are
$E=\left\{ H,H \right\},\left\{ H,T \right\},\left\{ T,H \right\}$.
We get the total number of favourable outcomes $=3$
So the probability of getting head at least once will be $P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$$\Rightarrow P\left( A \right)=\dfrac{3}{4}$

Note: The word “at least: means one or more than one. So, it is important to note while considering the possible outcomes we need to consider a case in which we are getting both heads. When we calculate the probability of occurrence of any event the value should be lie between 0-1. The value of probability cannot exceed one. Probability value higher than one means probability greater than $100\%$ and it is not possible.