Question

If you flip a fair coin $10$ times, what is the probability that it lands on heads exactly $4$ times?

Answer 1

In order to solve this question, we will first find out the probability of getting a head in a single toss. After that, using the formula of Binomial probability, we will find out the probability of getting exactly $4$ heads in total when a fair coin is tossed $10$ times.
Formulas used:
Probability of an event, $P\left( E \right) = \dfrac{{Favourable{\text{ }}outcomes}}{{Total{\text{ }}outcomes}}$
Binomial Probability is given by:
${}^n{C_r} \cdot {p^r} \cdot {\left( {1 - p} \right)^{n - r}}$
Here ${}^n{C_r}$ indicates the number of different combinations of $r$ objects from a set of $n$ objects.
$p$ is the probability of success on an individual trial.

Complete step by step answer:
It is given that a coin is tossed $10$ times
And we have to find the probability of getting exactly $4$ heads
Now first we will find out the probability of getting a head in a single toss,
We know that
When a single coin is tossed, then the outcomes are \left\\{ {H,T} \right\\}
So, total number of outcomes $= 2$
And favourable outcome (of getting a head) $= 1$
Now we know that,
Probability of an event, $P\left( E \right) = \dfrac{{Favourable{\text{ }}outcomes}}{{Total{\text{ }}outcomes}}$
Therefore, probability of getting a head in a single toss will be
$p = \dfrac{1}{2}$
Now using Binomial Theorem of Probability,
The probability of getting heads exactly $4$ times will be
${}^n{C_r} \cdot {p^r} \cdot {\left( {1 - p} \right)^{n - r}}$
Here, $n = 10,r = 4$
Therefore, the probability of getting exactly $r = 4$ heads in total $n = 10$ tosses, we get
${}^{10}{C_4} \cdot {p^4} \cdot {\left( {1 - p} \right)^{10 - 4}}$
$\Rightarrow {}^{10}{C_4} \cdot {p^4} \cdot {\left( {1 - p} \right)^6}$
On substituting the value of $p$ we get,
$\Rightarrow {}^{10}{C_4} \cdot {\left( {\dfrac{1}{2}} \right)^4} \cdot {\left( {1 - \dfrac{1}{2}} \right)^6}$
$\Rightarrow {}^{10}{C_4} \cdot {\left( {\dfrac{1}{2}} \right)^4} \cdot {\left( {\dfrac{1}{2}} \right)^6}$
We know that
${a^m} \times {a^n} = {a^{m + n}}$
Therefore, we get
$\Rightarrow {}^{10}{C_4} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
Now we know that
${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!{\text{ }} \cdot r!}}$
So, we get
$\Rightarrow \dfrac{{10!}}{{\left( {10 - 4} \right)!{\text{ }} \cdot 4!}} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
$\Rightarrow \dfrac{{10!}}{{6!{\text{ }} \cdot 4!}} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
Now, $10!$ can be written as $10 \times 9 \times 8 \times 7 \times 6!$
Therefore, we get
$\Rightarrow \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{6!{\text{ }} \cdot 4!}} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
On cancelling the terms, we get
$\Rightarrow \dfrac{{10 \times 9 \times 8 \times 7}}{{4!}} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
Now the value of $4! = 24$
Therefore, we get
$\Rightarrow \dfrac{{10 \times 9 \times 8 \times 7}}{{24}} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
On writing the factors of $10$ and $24$ we get
$\Rightarrow \dfrac{{2 \times 5 \times 9 \times 8 \times 7}}{{2 \times 2 \times 2 \times 3}} \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
On dividing the terms and cancelling the like terms, we get
$\Rightarrow 5 \times 3 \times 2 \times 7 \cdot {\left( {\dfrac{1}{2}} \right)^{10}}$
$\Rightarrow 5 \times 3 \times 7 \cdot {\left( {\dfrac{1}{2}} \right)^9}$
$\Rightarrow \dfrac{{5 \times 3 \times 7}}{{{2^9}}}$
On multiplying, we get
$\Rightarrow \dfrac{{105}}{{{2^9}}}$
which is the required answer.

Note:
In the question, the concept of Binomial probability is used because Binomial probability refers to the probability of exactly $r$ successes on $n$ repeated trials in an experiment which has two possible outcomes (commonly called a binomial experiment). Also note that if $p$ is the probability of success of a single trial, then $\left( {1 - p} \right)$ is the probability of failure of a single trial which is also represented by $q$ .Hence the formula of Binomial probability becomes ${}^n{C_r} \cdot {p^r} \cdot {q^{n - r}}$