Question

If a coin is tossed $n$ times, then the probability that the head comes odd times is
A. $\dfrac{1}{2}$
B. $\dfrac{1}{{{2}^{n}}}$
C. $\dfrac{1}{{{2}^{n-1}}}$
D. none of these

Answer 1

Hint: We first find the two events of conditional and unconditional events for probability. We break the conditional events in parts to find the number of ways using the binomial theorem. We also find the number of ways for an unconditional event \[n\left( S \right)\] that denotes the number of ways we can toss a coin to get a result. We find probability using \[p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\].

Complete step by step answer:
If a coin is tossed $n$ times, then the condition is that the head comes an odd number of times which means it will be 1, 3, 5, 7… and so on.
We individually take the number of ways each can be done.
For heads appearing 1 out of $n$ times, we get ${}^{n}{{C}_{1}}$.
For heads appearing 3 out of $n$ times, we get ${}^{n}{{C}_{3}}$.
For heads appearing 5 out of $n$ times, we get ${}^{n}{{C}_{5}}$.
Therefore, the total number of times is ${}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}+...$.
We have the binomial formula of \[{{\left( 1+x \right)}^{n}}=1+nx+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+....+{}^{n}{{C}_{n}}{{x}^{n}}\].
We replace $x$ with $-x$ to get \[{{\left( 1-x \right)}^{n}}=1-nx+{}^{n}{{C}_{2}}{{x}^{2}}-......+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{x}^{r}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{x}^{n}}\].
We subtract these two to get
\[\begin{align}
  & {{\left( 1+x \right)}^{n}}-{{\left( 1-x \right)}^{n}} \\
 & =\left( 1+nx+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+....+{}^{n}{{C}_{n}}{{x}^{n}} \right)-\left( 1-nx+{}^{n}{{C}_{2}}{{x}^{2}}-......+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{x}^{r}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{x}^{n}} \right) \\
 & =2\left( nx+{}^{n}{{C}_{3}}{{x}^{3}}+{}^{n}{{C}_{5}}{{x}^{5}}+......... \right) \\
\end{align}\]
We can write in simple form as
\[\begin{align}
  & nx+{}^{n}{{C}_{3}}{{x}^{3}}+{}^{n}{{C}_{5}}{{x}^{5}}+......... \\
 & ={}^{n}{{C}_{1}}x+{}^{n}{{C}_{3}}{{x}^{3}}+{}^{n}{{C}_{5}}{{x}^{5}}+......... \\
 & =\dfrac{{{\left( 1+x \right)}^{n}}-{{\left( 1-x \right)}^{n}}}{2} \\
\end{align}\]
We now put the value of $x=1$ to get
\[\begin{align}
  & {}^{n}{{C}_{1}}x+{}^{n}{{C}_{3}}{{x}^{3}}+{}^{n}{{C}_{5}}{{x}^{5}}+.........=\dfrac{{{\left( 1+x \right)}^{n}}-{{\left( 1-x \right)}^{n}}}{2} \\
 & \Rightarrow {}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}+.........=\dfrac{{{\left( 1+1 \right)}^{n}}-{{\left( 1-1 \right)}^{n}}}{2}=\dfrac{{{2}^{n}}}{2}={{2}^{n-1}} \\
\end{align}\]
We take this event as A where \[n\left( A \right)={{2}^{n-1}}\], the conditional event.
Now we take the universal event of \[n\left( S \right)\] where we find the unconditional event.
So, \[n\left( S \right)\] denotes the number of ways we can toss a coin to get a result. For each toss we have 2 results. So, \[n\left( S \right)={{2}^{n}}\].
The probability of A will be \[p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{{{2}^{n-1}}}{{{2}^{n}}}=\dfrac{1}{2}\].

So, the correct answer is “Option A”.

Note: We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact. That’s why we didn’t use the concept of number of points in a set and instead we directly used the numerical form to find the solution. To find individual probabilities, we divide them with $n\left( S \right)$.