Two unbiased coins are tossed simultaneously. Find the probability of getting :

(i)  at least one head

(ii)  atmost one head

(iii)  no head

(𝑖) 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 ℎ𝑒𝑎𝑑 

Let 𝐻 represent head and 𝑇 represent tails. 

𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇 

⇒ 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 4 

𝐹𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 𝐻𝐻, 𝐻𝑇, 𝑇𝐻 

⇒ 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 3

Therefore, the probability of getting ‘atleast one head’ is

$P\left(\text{ atleast one head }\right)=\frac{\text{ Number of favourable outcomes }}{\text{ Total Number of outcomes }}$

$\begin{array}{r}=\frac{3}{4}\\ \Rightarrow P\left(\text{ atleast one head }\right)=\frac{3}{4}\end{array}$

Therefore, the probability of getting “atleast one head” is $\frac{3}{4}$.

(𝑖𝑖) 𝑎𝑡𝑚𝑜𝑠𝑡 𝑜𝑛𝑒 ℎ𝑒𝑎𝑑 

𝐹𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 𝐻𝑇, 𝑇𝐻 

⇒ 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 2 

Therefore, the probability of getting ‘atmost one head’ is

$\begin{array}{r}P\left(\text{ atmost one head }\right)=\frac{\text{ Number of favourable outcomes }}{\text{ Total Number of outcomes }}\\ \Rightarrow P\left(\text{ atmost one head }\right)=\frac{2}{4}=\frac{1}{2}\end{array}$

(𝑖𝑖𝑖) 𝑛𝑜 ℎ𝑒𝑎𝑑 

𝐹𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 𝑇𝑇 

⇒ 𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 = 1 

Therefore, the probability of getting ‘no head’ is 

$\begin{array}{r}P\left(\text{ no head }\right)=\frac{\text{ Number of favourable outcomes }}{\text{ Total Number of outcomes }}\\ \Rightarrow P\left(\text{ no head }\right)=\frac{1}{4}\end{array}$