A fair die is thrown again and again. The probability that the second six is obtained on the fourth throw is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{25}{432}\)
The throws are Bernoulli trials with success = a six, so \(\displaystyle p = \frac{1}{6}\) and \(\displaystyle q = \frac{5}{6}\).
The second six comes on the fourth throw exactly when the first three throws contain exactly one six and the fourth throw is a six.
Exactly one six in the first three throws:
\( \displaystyle {}^{3}C_{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2} = \frac{75}{216} \)
A six on the fourth throw has probability \(\displaystyle\frac{1}{6}\), and the throws are independent:
\( \displaystyle P = \frac{75}{216} \times \frac{1}{6} = \frac{75}{1296} = \frac{25}{432} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{25}{72}\) — Incorrect. This is only the probability of exactly one six in the first three throws; the six on the fourth throw is left out.
Option (3) \(\displaystyle\frac{25}{216}\) — Incorrect. This is the probability of exactly two sixes in four throws, \({}^{4}C_{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{2}\), which also counts cases where the second six comes earlier.
Option (4) \(\displaystyle\frac{25}{1296}\) — Incorrect. This allows only one order (for example six, not, not, six); the first six can be on any of the first three throws.