Question

Determine the probability when a die is thrown 2 times such that there are no fours and no fives occur?

(a) \(\frac{4}{9}\)

(b) \(\frac{56}{89}\)

(c) \(\frac{13}{46}\)

(d) \(\frac{3}{97}\)

This question was addressed to me in examination.

My question comes from Counting in section Counting of Discrete Mathematics

Answer 1

Right option is (a) \(\frac{4}{9}\)

To explain: In this experiment, throwing a die anything other than a 4 is a success and rolling a 4 is failure. Since there are two trials, the required probability is

b(2; 2, \(\frac{5}{6}\)) = ^2C2 * (\(\frac{4}{6}\))^2 * (\(\frac{2}{6}\))^0 = \(\frac{4}{9}\).