Question

A fair coin tossed $n$ times. Let $X$ be the number of times head is observed. If $P\left( {X = 4} \right),P\left( {X = 5} \right),$ and $P\left( {X = 6} \right)$ are in H.P., then $n$ is equal to
1. $7$
2. $10$
3. $14$
4. None of these

Answer 1

Hint: In this question, we have three terms $P\left( {X = 4} \right),P\left( {X = 5} \right),$ and $P\left( {X = 6} \right)$ which are in Harmonic progression (H.P.) first convert them into A.P. as Harmonic Progression is a sequence of real numbers found by taking the arithmetic progression's reciprocals. Apply the condition of A.P. of three terms (the sum of its extremes is equal to twice the mid-term) and use the binomial formula. This question is lengthy so solve it till you can do it and put the value of $n$ which is given in the option then check whether both sides are equal. If not, it means the value is not true.

Formula Used:
Binomial distribution –
$P\left( {X = r} \right) = {}^n{C_r}{p^r}{q^{n - r}}$
Combination formula –
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right).......$

Complete step by step Solution:
Given that,
$P\left( {X = 4} \right),P\left( {X = 5} \right),$ and $P\left( {X = 6} \right)$ are in H.P where $X$ is the number of times head is observed when a fair coin is tossed $n$ times.
As we know, when a coin is tossed we get two outcomes either head or tail
Let the probability of getting head be successes,
$ \Rightarrow p = \dfrac{1}{2},q = 1 - p = \dfrac{1}{2}$
Now, applying formula of $r$ successes (Binomial distribution)
Therefore, $P\left( {X = r} \right) = {}^n{C_r}{p^r}{q^{n - r}} = {}^n{C_r}{\left( {\dfrac{1}{2}} \right)^n}$($\because p = q = \dfrac{1}{2}$)
As, $P\left( {X = 4} \right),P\left( {X = 5} \right),$ and $P\left( {X = 6} \right)$ are in H.P.
It implies that, $\dfrac{1}{{P\left( {X = 4} \right)}},\dfrac{1}{{P\left( {X = 5} \right)}},$ and $\dfrac{1}{{P\left( {X = 6} \right)}}$ are in A.P.
When three terms are in A.P. then the sum of its extremes is equal to twice the mid-term.
$\dfrac{1}{{2P\left( {X = 5} \right)}} = \dfrac{1}{{P\left( {X = 4} \right)}} + \dfrac{1}{{P\left( {X = 6} \right)}}$
$\dfrac{1}{{2P\left( {X = 5} \right)}} = \dfrac{{P\left( {X = 6} \right) + P\left( {X = 4} \right)}}{{P\left( {X = 4} \right) \times P\left( {X = 6} \right)}}$
$2P\left( {X = 5} \right) = \dfrac{{P\left( {X = 4} \right) \times P\left( {X = 6} \right)}}{{P\left( {X = 6} \right) + P\left( {X = 4} \right)}}$
$P\left( {X = 5} \right) = \dfrac{{2P\left( {X = 4} \right) \times P\left( {X = 6} \right)}}{{P\left( {X = 6} \right) + P\left( {X = 4} \right)}}$
Applying Binomial distribution,
${}^n{C_5}{\left( {\dfrac{1}{2}} \right)^n} = \dfrac{{2{}^n{C_4}{{\left( {\dfrac{1}{2}} \right)}^n}{}^n{C_6}{{\left( {\dfrac{1}{2}} \right)}^n}}}{{{}^n{C_4}{{\left( {\dfrac{1}{2}} \right)}^n} + {}^n{C_6}{{\left( {\dfrac{1}{2}} \right)}^n}}}$
Now, simplifying the above equation using factorial
$\dfrac{{n!}}{{5!\left( {n - 5} \right)!}}{\left( {\dfrac{1}{2}} \right)^n} = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^n}\left( {2{}^n{C_4}{}^n{C_6}} \right)}}{{{{\left( {\dfrac{1}{2}} \right)}^n}\left( {{}^n{C_4} + {}^n{C_6}} \right)}}$
$\dfrac{{n!}}{{5!\left( {n - 5} \right)!}}{\left( {\dfrac{1}{2}} \right)^n} = \dfrac{{\left( {2\dfrac{{n!}}{{4!\left( {n - 4} \right)!}} \times \dfrac{{n!}}{{6!\left( {n - 6} \right)!}}} \right)}}{{\left( {\dfrac{{n!}}{{4!\left( {n - 4} \right)!}} + \dfrac{{n!}}{{6!\left( {n - 6} \right)!}}} \right)}}$
$\dfrac{{n!}}{{5!\left( {n - 5} \right)!}}{\left( {\dfrac{1}{2}} \right)^n} = \left( {2\dfrac{{n!}}{{4!\left( {n - 4} \right)!}} \times \dfrac{{n!}}{{6!\left( {n - 6} \right)!}}} \right) \times \left( {\dfrac{{6!\left( {n - 6} \right)! \times 4!\left( {n - 4} \right)!}}{{n!6!\left( {n - 6} \right)! + n!4!\left( {n - 4} \right)!}}} \right)$
$\dfrac{1}{{5!\left( {n - 5} \right)!}}{\left( {\dfrac{1}{2}} \right)^n} = 2n! \times \left( {\dfrac{1}{{6!\left( {n - 6} \right)! + 4!\left( {n - 4} \right)!}}} \right)$
$\dfrac{1}{{5 \times 4!\left( {n - 5} \right)\left( {n - 6} \right)!}}{\left( {\dfrac{1}{2}} \right)^n} = 2n! \times \left( {\dfrac{1}{{6 \times 5 \times 4!\left( {n - 6} \right)! + 4!\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right)!}}} \right)$
$\dfrac{1}{{5\left( {n - 5} \right)}}{\left( {\dfrac{1}{2}} \right)^n} = 2n! \times \left( {\dfrac{1}{{6 \times 5 + \left( {n - 4} \right)\left( {n - 5} \right)}}} \right)$
Here, $\dfrac{1}{{5\left( {n - 5} \right)}}{\left( {\dfrac{1}{2}} \right)^n}$ is very small
At $n = 7,10,14$, both sides will not be equal
$ \Rightarrow n \ne 7,10,14$

Hence, the correct option is 4.

Note: The key concept involved in solving this problem is a good knowledge of H.P. (Harmonic progression). Students must know that to solve such questions first step is to convert H.P. into A.P. and to solve A.P. of three terms write the sum of its extremes equal to the twice of mid-term. Solve further using Binomial distribution.