A fair die is thrown again and again. The probability that a six appears for the first time on the third throw is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{25}{216}\)
The first six comes on the third throw exactly when the first throw is not a six, the second throw is not a six and the third throw is a six.
\( \displaystyle P(\text{not a six}) = \frac{5}{6},\quad P(\text{six}) = \frac{1}{6} \)
The throws are independent, so the three probabilities multiply:
\( \displaystyle P = \frac{5}{6}\times\frac{5}{6}\times\frac{1}{6} = \frac{25}{216} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{5}{36}\) — Incorrect. This is \(\frac{5}{6}\times\frac{1}{6}\), the chance that the first six comes on the second throw.
Option (3) \(\displaystyle\frac{1}{216}\) — Incorrect. This is the chance of a six on all three throws.
Option (4) \(\displaystyle\frac{1}{6}\) — Incorrect. This is only the chance of a six on the third throw; it leaves out the condition that the first two throws are not sixes.