Question

three unbiased coins are tossed simultaneously. Then the probability of receiving at the most two tails is ?

Answer 1

Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]

Favoured cases are = [TTH, THT, HTT, ] = 3

So required probability = 3/8 

Answer 2

Sample Space ={(H, H, H), (T,T,T),(H,H,T),(T,T,H),(H,T,H),(T,H,T),(H,T,T),(T,H,H)}
Number of favorable outcomes ={((T, T, H), (T, H, T), (H, T, T)}
Therefore, P(atmost two tails) =3/8.

HOPE THIS HELPS!!