Question

A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

Answer 1

Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12

(i) Let A be the event in which the sum of numbers that turn up is 3.
Accordingly, A = {(1, 2)}
$∴(P)(A)=\frac{\text{Number of outcomes favourable to A}}{\text{Total number of possible outcomes}}=\frac{n(A)}{n(S)}=\frac{1}{12}$

(ii) Let B be the event in which the sum of numbers that turn up is 12.
Accordingly, B = {(6, 6)}
$∴(P)(B)=\frac{\text{Number of outcomes favourable to B}}{\text{Total number of possible outcomes}}=\frac{n(A)}{n(B)}=\frac{1}{12}$