Question

If n coins are tossed then find the probability of not getting either all heads or all tails

• A. $1-\frac{1}{2^n}$
• B. $\frac{1}{2^{n-1}}$
• C. $1-\frac{1}{2^{n+1}}$
• D. $1-\frac{1}{2^{n-1}}$

Answer 1

The correct option is D $1-\frac{1}{2^{n-1}}$

If $n$ coins are tossed, $n\left(S\right)=2^n$.

Let $A$ be the event of getting all heads.

$\Rightarrow n\left(A\right)=1$

$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{2^n}$

Let $B$ be the event of getting all tails.

$\Rightarrow n\left(B\right)=1$

$\therefore P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{1}{2^n}$

Since $A$ and $B$ are mutually exclusive events, $A\cap B=\varphi$

Therefore, probability of getting either all heads or all tails is $P(A\cup B)=P\left(A\right)+P\left(B\right)$.

$\Rightarrow P(A\cup B)=\frac{1}{2^n}+\frac{1}{2^n}$

$=\frac{2}{2^n}$

$=\frac{1}{2^{n+-1}}$

$\therefore P(A\cup B)=\frac{1}{2^{n-1}}$

Now the probablity of not getting either all heads or all tails is given by $1-P(A\cup B)=1-\frac{1}{2^{n-1}}$.

Therefore, the probablity of not getting either all heads or all tails is $1-\frac{1}{2^{n-1}}$.