Question

A coin is tossed n times, what is the chance that the head will present itself an odd number of times?

Answer 1

In this solution, first we have to find the probability of one throw of a coin. Similarly, we will find the probability of two throws, three throws and n number of throws of a coin. Then, we will calculate the combination of the number of times the coin is thrown from which we will calculate the probability of the coins thrown.

Complete step-by-step answer:
Here,
We know that,
In one throw of a coin, the number of possible ways is $2$ since either head (H) or tail (T) may appear.
In two throws of a coin, the total number of ways will be $2 \times 2 = 4$

Since,
Corresponding to each way of the first coin there are $2$ ways of second.
Similarly,
In three throws of a coin, the total number of ways will be $2 \times 2 \times 2$ i,e $8$
Thus,
In n throws of a coin, the total number of ways will be ${\rm{N}} = {{\rm{2}}^{\rm{n}}}$
And,
M $=$ The favourable number of ways
So, the number of ways in which head will occur once or twice or $5$ times or $7$ times and so on,
$= {}^{\rm{n}}{{\rm{C}}_{\rm{1}}}{\rm{ + }}{}^{\rm{n}}{{\rm{C}}_{\rm{3}}}{\rm{ + }}{}^{\rm{n}}{{\rm{C}}_{{\rm{50}}}}{\rm{ + }}...... = {{\rm{2}}^{{\rm{n - 1}}}}$
Hence, the required probability will be,
$\begin{array}{l}{\rm{p}} = \dfrac{{\rm{M}}}{{\rm{N}}}\\\\\,\,\, = \dfrac{{{{\rm{2}}^{{\rm{n - 1}}}}}}{{{{\rm{2}}^{\rm{n}}}}}\\\\\,\,\, = \dfrac{{\rm{1}}}{{\rm{2}}}\end{array}$

Hence, the chance that the head will present itself an odd number of times will be $\dfrac{1}{2}$

Note: Probability explains the chances of happening in an event. Here we have to find the probability of the coins thrown. Since the total the combination of the number of times the coin is thrown can be calculated, so from that we can calculate the required probability of the coins thrown.