Question

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

Answer 1

Hint: 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 \[{}^n{C_r} \cdot {p^r} \cdot {q^{n - r}}\]