When three coins are tossed simultaneously, the sample space becomes 23 = 8
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Probability of getting at most two heads:
Let A be the event of getting at most two heads.
So, the favourable outcomes are {HHT,HTH,TTT,THH,TTH,THT,HTT}
Therefore, n(A) = 7
Hence, the probability of getting at most two heads = Number of favourable outcomes/Total number of outcomes = ⅞.
Probability of getting at most two tails:
Let B be the event of getting at most two tails.
So, the favourable outcomes are {HHH, HHT, HTH, THH, TTH, THT, HTT}
Thus, n(B) = 7
Hence, the probability of getting at most two tails = ⅞.