A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At least two heads are observed } C:{ The number of heads observed is odd } Find the following probabilities by summing the probabilities of the appropriate sample points (note that 0 is an even number): (a) P(B) =(b) P(A or B) = (c) P(A or B or C)

Answer:(a) 1/2(b) 1/2(c) 1/8Step-by-step explanation:Since, when a fair coin is tossed three times,The the total number of possible outcomesn(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,[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex](a) So, the probability of B,[tex]P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}[/tex](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,[tex]\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}[/tex](c) C: { The number of heads observed is odd },⇒ C = { HHH, HTT, THT, TTH},∵ A ∩ B ∩ C = {HHH},⇒ n(A ∩ B ∩ C) = 1,[tex]\implies P(A\cap B\cap C)=\frac{1}{8}[/tex]