Question

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to that of getting 9 heads, then the probability of 3 heads is
A. $\dfrac{35}{{{2}^{12}}}$
B. $\dfrac{35}{{{2}^{14}}}$
C. $\dfrac{7}{{{2}^{12}}}$
D. none of these

Answer 1

Hint: At first we will explain the concept of empirical probability and how the events are considered. We take the given events and find the number of outcomes. Using the probability theorem of $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$, we get the empirical probability of the events and solve the equation.

Complete answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome.
We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options.
We assume that the fair coin is tossed $n$ a number of times. We take the conditional event A as getting 7 heads, conditional event B as getting 9 heads and the universal event U as tossing $n$ number of times and numbers will be denoted as $n\left( A \right)$, $n\left( B \right)$, $n\left( U \right)$ respectively.
As the probabilities are the same, the number of outcomes will also be the same.
We get that $n\left( A \right)={}^{n}{{C}_{7}}$, $n\left( B \right)={}^{n}{{C}_{9}}$, $n\left( A \right)=n\left( B \right)$.
So, the condition of ${}^{n}{{C}_{9}}={}^{n}{{C}_{7}}$ gives $n=9+7=16$.
 We now assume the conditional event C as getting 3 heads and the number is $n\left( C \right)$.
So, $n\left( C \right)={}^{16}{{C}_{3}}$ and $n\left( U \right)={{2}^{n}}={{2}^{16}}$.
We take the empirical probability of the given problem as $P\left( C \right)=\dfrac{n\left( C \right)}{n\left( U \right)}$.
The empirical probability of the shooting event is $P\left( C \right)=\dfrac{{}^{16}{{C}_{3}}}{{{2}^{16}}}=\dfrac{16\times 15\times 14}{{{2}^{16}}\times 3!}=\dfrac{35}{{{2}^{12}}}$.
And hence the correct answer is option A.

Note:
We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.