Three identical coins are tossed together. What is the probability of obtaining :

(i) all heads ?

(ii) exactly two heads ?

(iii) exactly one head ?

(iv) no head ?

When three identical coins are tossed together, the total number of possible outcomes = 8 (i.e. HHH, HHT, HTH, THH, TTH, THT, HTT and TTT)

(i) Number of favourable outcomes (Getting all heads) = 1 (HHH)

P(Getting all heads) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{1}{8}$

Hence, the probability of getting all heads is $\dfrac{1}{8}$.

(ii) Number of favourable outcomes (Getting exactly two heads) = 3 (HHT, HTH, THH)

P(Getting exactly two heads) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{3}{8}$

Hence, the probability of getting exactly two heads is $\dfrac{3}{8}$.

(iii) Number of favourable outcomes (Getting exactly one head) = 3 (HTT, TTH, THT)

P(Getting exactly one head) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{3}{8}$

Hence, the probability of getting exactly one head is $\dfrac{3}{8}$.

(iv) Number of favourable outcomes (Getting no head) = 1 (TTT)

P(Getting no head) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{1}{8}$

Hence, the probability of getting exactly no head is $\dfrac{1}{8}$.