Question

A fair coin is tossed 8 times. Find the probability that it shows heads, exactly 5 times.

Answer 1

Hint: In this question we first need to find the probability of getting a head and tail when a coin is tossed using the formula \[P=\dfrac{m}{n}\]. Then we need to use the formula of Bernoulli trial to get the probability of getting 5 heads which can be found by using the formula \[P\left( X \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\]. Now, on substituting the respective values and simplifying further we get the result.

Complete step by step solution:
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\]
Bernoulli Trial:
In a random experiment, if there are any two events, "Success and Failure" and the sum of the probabilities of these two is one, then any outcome of such experiment is known as a Bernoulli trial.
The probability of r successes in n independent Bernoulli trials is given by
\[P\left( X \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\]
Where, p is the probability of success and q is the probability of failure.
Now, when a coin is tossed then there is a possibility of two outcomes which can be either head or tail.
\[m=1,n=2\]
Here, let us assume that the event of getting a head as p and event of getting a tail as q.
Now, the probability of getting a head is given by
\[\Rightarrow P=\dfrac{m}{n}\]
Now, on substituting the respective values we get,
\[\Rightarrow p=\dfrac{1}{2}\]
Let us now find the probability of getting a tail
\[m=1,n=2\]
Now, on substituting these values in the probability formula we get,
\[\Rightarrow q=\dfrac{1}{2}\]
Now, we need to find the probability of getting 5 heads when a coin is tossed 8 times
\[P\left( X \right)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}\]
Now, from the given conditions in the question we have
\[n=8,r=5,p=\dfrac{1}{2},q=\dfrac{1}{2}\]
Now, on substituting these values in the above Bernoulli formula we get,
\[\Rightarrow P\left( X \right)={}^{8}{{C}_{5}}{{\left( \dfrac{1}{2} \right)}^{5}}{{\left( \dfrac{1}{2} \right)}^{8-5}}\]
Now, this can be further written as
\[\Rightarrow P\left( X \right)=\dfrac{8!}{5!3!}{{\left( \dfrac{1}{2} \right)}^{8}}\]
Now, on further simplification we get,
\[\Rightarrow P\left( X \right)=\dfrac{8\times 7\times 6}{6}\times {{\left( \dfrac{1}{2} \right)}^{8}}\]
Now, on cancelling the common terms and simplifying it further we get,
\[\Rightarrow P\left( X \right)=\dfrac{7}{{{2}^{5}}}\]
Now, on simplifying this further we get,
\[\therefore P\left( X \right)=\dfrac{7}{32}\]

Note: Instead of considering the Bernoulli trial to simplify this we can also solve it by considering all the possible outcomes when a coin is tossed and then further check the total number of outcomes having exactly 5 heads and then substitute that in the probability formula.
It is important to note that a given coin is fair so the sum of probabilities of head and tail will be 1 so that we can use the Bernoulli trial to simplify further. Here, while substituting the values we need to substitute the respective ones and then simplify further accordingly.