Question

A coin is tossed $n$ times. The chance that the head will present itself an odd number of times is $\dfrac{1}{k}$ . find the value of $k$

Answer 1

To find the probability of the chance of the head presenting itself an odd number of times we need to find out the total number of possible outcomes and find the number of ways the head is showing itself. Next find the ratio of these two outcomes. This will give the chance of heads showing out of $n$ number of times the coin is tossed.

Complete step-by-step solution:
Given the coin is tossed $n$ times.
The outcomes of tossing coins one time are either head or tail i.e., 2 outcomes.
The total number of outcomes are for tossing $n$ times is given by
$2\times 2\times 2......n\,times={{2}^{n}}$
Now consider the total outcomes of the head showing an odd number of terms.
If coin shows head once then the total outcomes of head showing odd number of terms is given by ${}^{n}{{C}_{1}}$ and for three times it is given by ${}^{n}{{C}_{3}}$
Similarly, when we add these outcomes, it gives the total outcomes of the head showing an odd number of times.
The equation will be
${}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}+{}^{n}{{C}_{7}}.....$
This sum will be equal to ${{2}^{n-1}}$
Hence, we get ${}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}+{}^{n}{{C}_{7}}.....={{2}^{n-1}}$
Now the chances of head present odd number of times are given by $\dfrac{1}{k}$
The formula for the chances of head present on the odd number of times
$=\dfrac{\text{total outcomes of head showing odd number of times}}{\text{total outcomes of coin tossed}}$
$\Rightarrow \dfrac{1}{k}=\dfrac{\text{total outcomes of head showing odd number of times}}{\text{total outcomes of coin tossed}}$
Substitute the values
$\Rightarrow \dfrac{1}{k}=\dfrac{{{2}^{n-1}}}{{{2}^{n}}}$
$\Rightarrow \dfrac{1}{k}=\dfrac{{{2}^{n}}\times {{2}^{-1}}}{{{2}^{n}}}$
$\Rightarrow \dfrac{1}{k}=\dfrac{1}{2}$
As the LHS side is equal to RHS the value of $k$ will be equal to $2$.
$k=2$

Note: The addition of combinations will be given by ${}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}.....={{2}^{n}}$
When we consider odd terms only we have ${}^{n}{{C}_{1}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{5}}+{}^{n}{{C}_{7}}.....={{2}^{n-1}}$