In throwing coin and dice obtained sample space
S = {H, T} × {1, 2, 3, 4, 5, 6}
= {H1, H2, H3, H4, H5, H6, T1, T2, 73, T4, T5, T6}
n(S) = 12
Let A be the sample space of getting tail on coin and event number on dice, then
A = {H2, H4, H6}
n(A) = 3
Thus, required probability
P(A) = n(A)/n(S)
P(A) = 3/12 = 1/4
Thus, required probability is 1/4.