Question

In a simultaneous toss of two coins, find the probability of getting:
(i)2 heads (ii) Exactly one tail (iii) No tails

Answer 1

When two different coins are tossed simultaneously, the sample space is given by

$S=\{HH,HT,TH,TT\}$

Therefore, $n\left(S\right)=4$.

(1) Getting two heads:

Let $E_1$ be an event of getting 2 heads.

Then, $E_1=\left\{HH\right\}$ and, therefore, $n\left(E_1\right)=1$.

Therefore, P(getting 2 heads) $=P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{1}{4}$.

(2) Getting one tail:

Let $E_2$ = event of getting 1 tail. Then,

$E_2=\{TH,HT\}$ and, therefore, $n\left(E_2\right)=2$.

Therefore, P(getting 1 tail) $=P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{2}{4}=0.5$

(3) Getting no tail:

Let $E_3$ = event of getting no tail. Then,

$E_3=\left\{HH\right\}$ and, therefore, $n\left(E_3\right)=1$.

Therefore, P(getting no tail) $=P\left(E_3\right)=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{1}{4}$