A die is thrown thrice. If the first throw is a four, then the chance of getting 11 as total sum, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{6} \)
The first throw is known to be 4, so only the last two throws matter, and their sum must be 11 − 4 = 7.
The pairs with sum 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1): 6 out of 36 equally likely pairs.
\( \displaystyle P=\frac{6}{36} \)
\( \displaystyle =\frac{1}{6} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{6}{216} \) — Incorrect. This is the probability that the first throw is 4 and the total is 11; the question says the first throw is already a four.
Option (2) \( \displaystyle\frac{1}{216} \) — Incorrect. The probability is \( \frac{1}{6} \), not \( \frac{1}{216} \).
Option (4) \( \displaystyle\frac{7}{36} \) — Incorrect. Only 6 of the 36 pairs give the sum 7, so the probability is \( \frac{6}{36} \), not \( \frac{7}{36} \).