Three coins are tossed. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads?


Solution:

Let ‘S’ be the sample – space. Then S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }
(i) Let ‘E1’ = Event of getting all heads, Then E1 = { HHH }
 |E1| = 1
 P(E1) = |E1| / |S| = 1 / 8
(ii) Let E2 = Event of getting ‘2’ heads. Then:
 E2 = { HHT, HTH, THH }
 |E2| = 3
 P (E2) = 3 / 8
(iii) Let E3 = Event of getting at least one head. Then:
 E3 = { HHH, HHT, HTH, THH, HTT, THT, TTH }
 |E3| = 7
 P (E3) = 7 / 8
(iv) Let E4 = Event of getting at least one head, Then:
 E4 = { HHH, HHT, HTH, THH, }
 |E4| = 4
 P (E4) = 4/8 = 1/2