Question

A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the number of times one gets tail is
A. $ \dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}.{\left( {\dfrac{1}{2}} \right)^2}n $
B. $ 1 - \dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}} $
C. $ 1 - \dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}.\dfrac{1}{{{4^n}}} $
D. None of these

Answer 1

Hint: Probability of an event is the ratio of Number of favorable outcomes and total number of outcomes. So note down the favorable outcomes of the event when a coin is tossed as per the conditions and total number of outcomes when the coin is tossed. When a coin is tossed 2n times then the total number of outcomes is $ {2^{2n}} $

Complete step-by-step answer:
We are given that a coin is tossed 2n times.
We have to find the probability the number of times one gets head is not equal to the number of times one gets tail.
When a coin is tossed 2n times, the no. of outcomes possible are $ {2^{2n}} $
The probability among all the events is distributed as $ \Rightarrow \dfrac{{{}^{2n}{C_0}}}{{{2^{2n}}}},\dfrac{{{}^{2n}{C_1}}}{{{2^{2n}}}},\dfrac{{{}^{2n}{C_2}}}{{{2^{2n}}}},.....,\dfrac{{{}^{2n}{C_{2n - 1}}}}{{{2^{2n}}}},\dfrac{{{}^{2n}{C_{2n}}}}{{{2^{2n}}}} $
The sum of the probabilities of all the events is equal to 1.
 $
\Rightarrow \dfrac{{{}^{2n}{C_0}}}{{{2^{2n}}}} + \dfrac{{{}^{2n}{C_1}}}{{{2^{2n}}}} + \dfrac{{{}^{2n}{C_2}}}{{{2^{2n}}}} + ..... + \dfrac{{{}^{2n}{C_{2n - 1}}}}{{{2^{2n}}}} + \dfrac{{{}^{2n}{C_{2n}}}}{{{2^{2n}}}} = 1 \\
 \Rightarrow \dfrac{1}{{{2^{2n}}}}\left( {{}^{2n}{C_0} + {}^{2n}{C_1} + {}^{2n}{C_2} + ...... + \Rightarrow {}^{2n}{C_{2n}}} \right) = 1 \\
  $
And the probability of getting r heads is $ P\left( r \right) = \dfrac{1}{{{2^{2n}}}}{}^{2n}{C_r} $
Probability of getting the same number of heads and tails is the middle term of the probability distribution.
The middle term is
  $
\Rightarrow P\left( n \right) = \dfrac{1}{{{2^{2n}}}}{}^{2n}{C_n} \\
\Rightarrow {2^{2n}} = {4^n} \\
\Rightarrow {}^{2n}{C_n} = \dfrac{{\left( {2n} \right)!}}{{n!\left( {2n - n} \right)!}} = \Rightarrow \dfrac{{\left( {2n} \right)!}}{{n!.n!}} = \dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}} \\
\Rightarrow P\left( n \right) = \dfrac{1}{{{4^n}}}.\dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}} \\
  $
Probability of not getting the same number of heads as tails is 1 - Probability of getting the same number of heads and tails (Complementary events).
 $
\Rightarrow P\left( {{n^1}} \right) = 1 - \left( {\dfrac{1}{{{4^n}}} \times \dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}} \right) \\
  \therefore P\left( {{n^1}} \right) = 1 - \dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}} \cdot \dfrac{1}{{{4^n}}} \\
 $
The correct option is Option C.

So, the correct answer is “Option C”.

Note: Two events are considered to be complementary, when one event occurs if and only if the other event does not occur. The probabilities of the complementary events sum up to 1.
We know that the expansion of $ {\left( {1 + x} \right)^n} $ using binomial expansion is $ {}^n{C_0}{x^0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + .... + {}^n{C_n}{x^n} $
When the value of x is 1 and n is 2n, then
  $
 \Rightarrow {\left( {1 + 1} \right)^n} = {}^n{C_0}{1^0} + {}^n{C_1}{1^1} + {}^n{C_2}{1^2} + .... + {}^n{C_n}{1^n} \\
  \Rightarrow {2^n} = {}^n{C_0}1 + {}^n{C_1}1 + {}^n{C_2}1 + .... + {}^n{C_n}1 \\
  \Rightarrow {2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ..... + {}^n{C_{n - 1}} + {}^n{C_n} \\
 \Rightarrow {2^{2n}} = {}^{2n}{C_0} + {}^{2n}{C_1} + {}^{2n}{C_2} + ..... + {}^{2n}{C_{2n - 1}} + {}^{2n}{C_{2n}} \\
  $