Question

A die is thrown twice. What is the probability that: (i) 5 will not come up either time? (ii) 5 will come up at least once

Answer 1

Total number of outcomes = $6\times 6=36$
(i) Total number of outcomes when 5 comes up on either time are (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)
Hence, total number of favourable cases = 11
P (5 will come up either time)
= $\frac{11}{36}$
P (5 will not come up either time)
= $1-\frac{11}{36}$
= $\frac{25}{36}$
(ii) Total number of cases, when 5 can come at least once = 11
P (5 will come at least once)
= $\frac{11}{36}$