Two unbiased coins are tossed simultaneously. Find the probability of getting at most 1 head?

Hint: In this question, we use simple probability formula for favorable events.
Probability \[=\dfrac{favorable\text{ }cases}{total\text{ }possible\text{ }cases}\]
We count all cases of tossing two coins simultaneously and then calculate probability. We will use the fact that when one coin is tossed, the outcome can be head or tail, similarly, we can find total outcomes when 2 coins are tossed.

Complete step by step answer:

Now let us start with understanding the concept first. If there are $'n'$ outcomes in an experiment, which is equally likely, mutually exclusive, and exhaustive, and out of $'n'$ outcomes, $'m'$outcomes are in favor of happening an event A, then the probability of happening of event A –
\[p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{m}{n}\].
According to the question, we have an experiment of tossing two unbiased coins simultaneously.
We know that on tossing an unbiased coin, there are two outcomes – ‘Head and Tail’. So on tossing two unbiased coins, there are $4$ outcomes –
(a) Head, Head
(b) Head, Tail
(c) Tail, Head
(d) Tail, Tail

So let total possible outcomes of experiment is equal to$n$, and
$n\left( S \right)=n=4$.
Now we have to calculate the probability of event A, i.e., getting at most one head in the experiment. This means, either we get zero head or we get one head, but head in both coins in not favorable for event A.
 So cases are –
(1) Tail, Tail $\to $ zero head
(2) Tail, Head $\to $ one head
(3) Head, Tail $\to $ one head

So, number of favorable cases
$n\left( A \right)=m=3$
According to formula of probability,
\[p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{m}{n}\]
\[\Rightarrow p\left( A \right)=\dfrac{3}{4}\]
\[\Rightarrow p\left( A \right)=0.75\]

Hence, the probability of getting at most one head in the experiment of tossing two coins simultaneously is$0.75$ or\[\dfrac{3}{4}\].


Note:
In this question, we can get favorable cases by the following the formula also –
Number of getting at most one head $=$ total possible cases$-$number if getting more than one head.
So,
$n\left( A \right)=4-no.\text{ }of\text{ }cases\text{ }of\text{ }getting\text{ }tail\text{ }in\text{ }both\text{ }coins$
\[\Rightarrow n\left( A \right)=4-1\]
\[\Rightarrow n\left( A \right)=3\], and$n\left( S \right)=4$.

So, probability \[p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]
\[=\dfrac{m}{n}\]
\[=\dfrac{3}{4}\] .

(ii) In this type of question when we are asked about more than two coins tossed, then we may get forget some cases. So, students should keep in mind this and never forget any case. Else, this will make your whole answer wrong.