Answer:
(a) 1/2
(b) 1/2
(c) 1/8
Step-by-step explanation:
Since, when a fair coin is tossed three times,
The the total number of possible outcomes
n(S) = 2 × 2 × 2
= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },
Here, B : { At least two heads are observed } ,
⇒ B = {HHH, HHT, HTH, THH},
⇒ n(B) = 4,
Since,

(a) So, the probability of B,

(b) A : { At least one head is observed },
⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},
∵ A ∩ B = {HHH, HHT, HTH, THH},
n(A∩ B) = 4,

(c) C: { The number of heads observed is odd },
⇒ C = { HHH, HTT, THT, TTH},
∵ A ∩ B ∩ C = {HHH},
⇒ n(A ∩ B ∩ C) = 1,
