Question

If n coins are tossed simultaneously, the probability of getting head an odd number of times is
A) $ \dfrac{1}{{{2^n}}} $
B) $ \dfrac{1}{3} $
C) $ \dfrac{1}{2} $
D) $ \dfrac{1}{{{2^{n - 1}}}} $

Answer 1

Hint: This is a probability question. First we will derive the total number of ways and the number of ways of getting head odd times. Then we will get the probability by dividing the number of favorable ways with the total number of ways.

Complete step-by-step answer:
The number of possible ways in one toss of a coin is 2 i.e. head(H) or tail(T) might appear.
In two tosses of a coin, the total number of possible ways is $ 2 \times 2 $ i.e. $ {2^2} $ since corresponding to each way of the first toss of the coin there were two ways.
Thus for n number of toss, the number of possible ways $ = 2 \times 2 \times 2 \times 2.............. $ n times
Hence, the number of possible ways for n number of toss is $ {2^n} $
The number of possible ways in which the head will occur is odd times i.e. once, thrice, 5 times, 7 times and so on.
The number of possible ways in which head will occur is $ = {}^n{C_1} + {}^n{C_3} + {}^n{C_5} + ........... + {}^n{C_n} $ (let If n is odd.)
We know that the sum of the coefficient of expansion,
 $ {(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n} $ ……………………….. (1)
Hence, the sum of the coefficient of expansion $ {2^n} $ ,
 $ {2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ....... + {}^n{C_n} $ ………………………. (2)
Putting x = -1 in equation 1 we get,
 $ 0 = {}^n{C_0} - {}^n{C_1} + {}^n{C_2} - ....... + {}^n{C_n} $ …………………….(3)
Subtracting equation 3 from 2 we get,
 $ {2^n} = 2({}^n{C_1} + {}^n{C_3} + {}^n{C_5} - ....... + {}^n{C_n}) $
Expanding it we get,
 $ ({}^n{C_1} + {}^n{C_3} + {}^n{C_5} - ....... + {}^n{C_n}) = \dfrac{{{2^n}}}{2} = {2^{n - 1}} $
The number of possible ways in which head will occur is $ {2^{n - 1}} $
Probability of getting head = number of favorable ways / total number of ways
 $ = \dfrac{{{2^{n - 1}}}}{{{2^n}}} = \dfrac{1}{2} $
Hence, the probability of getting a head an odd number of times is \[\dfrac{1}{2}\].

Option C is correct.

Note: Probability is concerned with numerical description of how likely an event is to occur and how likely it is that a proposition is true. It is always between 0 and 1.
The formula of probability in simple language is, number of favorable outcomes/total outcomes.
You should practice similar questions of probability related to cards, dice, coins, balls etc.
You might make mistakes while deriving equations. So, be cautious while doing that.