Question

Solve the following question and mark the best possible option.
Let a, b, c be the respective numbers that show up when three fair dice, P1, P2, and P3, are rolled. If z = |(a – 1) (b – 3) (c – 4)|,the probability that z ≤ 1 is

Options

• `109/216`

• `95/216`

• `97/216`

• `93/216`

Answer 1

z ≤ 1 ⇒ z = 0 or z = 1.
Consider z = 0 ⇒ |(a – 1) (b – 3) (c – 4)| = 0 ⇒ a = 1, b = 3, and c = 4.
Probability that a = 1 or b = 3 or c = 4 = 1 - Probability that a ≠ 1, b ≠ 3 or c ≠ 4.
= 1 - `(5/6)(5/6)(5/6) = (91/216)`

Consider z = 1 ⇒ |(a – 1) (b – 3) (c – 4)| = 1
                        ⇒ a = 2, b = 2 or 4, c = 3 or 5.

Probability that a = 2 and b = 2 or 4 and c = 3 or 5 is `1/6 xx 2/6 xx 2/6 = 4/216`

∴ Probability that  z = 0 or z = 1 is `91/216 + 4/216 = 95/216`.