Question

A fair coin is tossed $n$ times. If the probability that head occurs 6 times is equal to the probability that head occurs 8 times, then $n$ is equal to
A. 15
B. 14
C. 12
D. 7

Answer 1

Hint:Here the given question is based on the concept of probability. We have to find the number of times which coin tossed on equating the given condition i.e., probability of head occurs 6 times to the probability of head occurs 8 times by Bernoulli condition $P\left( {X = x} \right){ = ^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$, and further simplify by using a combination formula to get the required solution.

Complete step by step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event. The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
$\text{Probability of event to happen}\,P\left( E \right) = \dfrac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}$

Consider the question: A fair coin is tossed $n$ times, we need to find the value of $n$.
The number of trails $n = ?$
Let us take ‘$p$’ be the probability of head and ‘$q$’ be the probability of tail.
Probability of head $p = \dfrac{1}{2}$ ----(1)
Probability of tail $q = \dfrac{1}{2}$ ------(2)
Given the condition, the probability of that head occurs 6 times is equal to the probability of that head occurs 8 times i.e., we have,
$P\left( {X = 6} \right) = P\left( {X = 8} \right)$ -------(1)
Let $X$ be the random variable that represents the number of successes in four trials. It’s a Bernoulli trial so, $X$ has a binomial distribution, we obtain
$P\left( {X = x} \right){ = ^n}{C_x}{\left( p \right)^x}{\left( q \right)^{n - x}}$
Then equation (1) becomes
$ \Rightarrow \,\,\,{\,^n}{C_6}{\left( {\dfrac{1}{2}} \right)^6}{\left( {\dfrac{1}{2}} \right)^{n - 6}} = \,\,{\,^n}{C_8}{\left( {\dfrac{1}{2}} \right)^8}{\left( {\dfrac{1}{2}} \right)^{n - 8}}$

We know the property of exponent ${a^m}{a^n} = {a^{m + n}}$, then
$ \Rightarrow \,\,\,{\,^n}{C_6}{\left( {\dfrac{1}{2}} \right)^{6 + n - 6}} = \,\,{\,^n}{C_8}{\left( {\dfrac{1}{2}} \right)^{8 + n - 8}}$
$ \Rightarrow \,\,\,{\,^n}{C_6}{\left( {\dfrac{1}{2}} \right)^n} = \,\,{\,^n}{C_8}{\left( {\dfrac{1}{2}} \right)^n}$
On cancelling the like terms ${\left( {\dfrac{1}{2}} \right)^n}$ on both side
$ \Rightarrow \,\,\,{\,^n}{C_6} = {\,^n}{C_8}$
We know the formula of combination $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, then
$ \Rightarrow \,\,\,\,\dfrac{{n!}}{{\left( {n - 6} \right)!6!}} = \dfrac{{n!}}{{\left( {n - 8} \right)!8!}}$
Again, on cancelling the like terms $n!$ on both side in numerator, then
$ \Rightarrow \,\,\,\,\dfrac{1}{{\left( {n - 6} \right)!6!}} = \dfrac{1}{{\left( {n - 8} \right)!8!}}$
On cross multiplication
$ \Rightarrow \,\,\,\,\dfrac{{\left( {n - 6} \right)!}}{{\left( {n - 8} \right)!}} = \dfrac{{8!}}{{6!}}$
$ \Rightarrow \,\,\,\,\dfrac{{\left( {n - 6} \right)\left( {n - 7} \right)\left( {n - 8} \right)!}}{{\left( {n - 8} \right)!}} = \dfrac{{8 \times 7 \times 6!}}{{6!}}$

On simplification, we get
$ \Rightarrow \,\,\,\,\left( {n - 6} \right)\left( {n - 7} \right) = 56$
$ \Rightarrow \,\,\,\,{n^2} - 7n - 6n + 42 = 56$
$ \Rightarrow \,\,\,\,{n^2} - 13n + 42 = 56$
Subtract 56 on both side
$ \Rightarrow \,\,\,\,{n^2} - 13n + 42 - 56 = 0$
$ \Rightarrow \,\,\,\,{n^2} - 13n - 14 = 0$
Apply a method of factorization, then
$ \Rightarrow \,\,\,\,{n^2} - 14n + n - 14 = 0$
$ \Rightarrow \,\,\,\,{n^2} - 14n + n - 14 = 0$
$ \Rightarrow \,\,\,\,n\left( {n - 14} \right) + 1\left( {n - 14} \right) = 0$
$ \Rightarrow \,\,\,\,\left( {n - 14} \right)\left( {n + 1} \right) = 0$
$ \Rightarrow \,\,\,\,n - 14 = 0$ or $n + 1 = 0$
$\therefore \,\,\,\,n = 14$ or $n = - 1$
The number of coins tossed never be in negative numbers. Hence the required value of $\,\,n = 14$.

Therefore, option B is the correct answer.

Note:The probability is a number of possible values. Candidates must know that we have to use the permutation concept or combination concept to solve the given problem because it is the first and main thing to solve the problem. Remember, factorial is the continued product of first n natural numbers is called the “n factorial” and it is represented by $n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot 3 \cdot 2 \cdot 1$.