Q26: Suppose three dice are rolled. Under the condition that no... - Statistical Methods & Experimental Designs - ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper | Agrigyan

Mathematical calculation of conditional probability:
Let event $B$ be 'no two dice show the same face' (all three dice show distinct numbers from 1 to 6).
Let event $A$ be 'one of the faces is an ace (number 1)'.

• Incorrect: $1/4$ underestimates the favorable outcomes.
• Correct:
  Total number of permutations where all 3 dice show distinct faces:
  $$n(B) = 6 \times 5 \times 4 = 120$$
  Number of permutations where all 3 faces are distinct AND one of them is an ace (1):
  The ace can appear on any of the 3 dice ($3$ choices). The remaining two dice must take distinct values chosen from the other 5 numbers ($5 \times 4 = 20$ choices):
  $$n(A \cap B) = 3 \times (5 \times 4) = 60$$