Answer:
b. S = AUB
Step-by-step explanation:
Since the coins are tossed 3 times and each coin has head, H and tail, T(2 sides), the sample space is S = 2 × 2 × 2 = 2³ = 8
All the possible outcomes are HTT, HHT, HHH, THH, TTH, HTH,THT and TTT
Since S denote the sample space
S = {HTT, HHT, HHH, THH, TTH, HTH,THT, TTT}
Since A denote the event that at least 1 of the coins comes to rest with its heads side upwards, the possible outcomes are HTT, HHT, HHH, THH, TTH, HTH and THT
So, A = {HTT, HHT, HHH, THH, TTH, HTH,THT}
Also B denote the event that none of the coins comes to rest with its heads side upwards, that is no heads. The possible outcome is TTT
So, B = {TTT}
Since S denote the sample space
S = {HTT, HHT, HHH, THH, TTH, HTH,THT, TTT}
So, A ∪ B = {HTT, HHT, HHH, THH, TTH, HTH,THT} ∪ {TTT} = {HTT, HHT, HHH, THH, TTH, HTH,THT, TTT} = S
So, S = A ∪ B
So, S = A ∪ B does not denote an abuse of notation.
The answer is b.
