Given that A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears and We need to find What is the probability that a number greater than 4 will first appear on the third or fourth roll?

At each stage The Probability of Getting a number Greater than 4 = \(\frac{2}{6}\) (As there are two numbers greater than 4 i.e. 5 and 6 out of 1 to 6) = \(\frac{1}{3}\)

Similarly, Probability of Getting a number less than or equal to 4 = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll)

P(number greater than 4 will first appear on the third roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{1}{3}\) = \(\frac{4}{27}\)

P(number greater than 4 will first appear on the fourth roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give ≤ 4) * P(Fourth Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) *\(\frac{1}{3}\) = \(\frac{8}{81}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll) = \(\frac{4}{27}\) + \(\frac{8}{81}\) = \(\frac{(12 + 8)}{81}\) = \(\frac{20}{81}\)

So, Answer will be E
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems