A die is rolled three times. The probability of getting a larger number than the previous number each time is -

Correct — option (4), 5/54.

Step 1 — count all outcomes.
Each roll shows 1 to 6.
6 × 6 × 6 = 216 equally likely outcomes

Step 2 — count the favourable outcomes.
"A larger number than the previous number each time" means first < second < third.
Any 3 different numbers from 1 to 6 can be rolled in increasing order in exactly one way.
Favourable = C(6, 3) = (6 × 5 × 4) ÷ (3 × 2 × 1) = 20

Step 3 — divide.
P = 20 ÷ 216 = 5/54

Check by listing, grouped by the first number:
First roll 1: choose 2 of 2–6, C(5, 2) = 10
First roll 2: choose 2 of 3–6, C(4, 2) = 6

First roll 3: choose 2 of 4–6, C(3, 2) = 3
First roll 4: only 4, 5, 6 = 1
Total = 10 + 6 + 3 + 1 = 20

The idea to remember: a strictly increasing result is fixed by which numbers appear, so count selections, C(6, 3), not arrangements.

Why the others are wrong

• (1)
  5/72 — 5/72 = 15/216, which would mean 15 favourable outcomes. There are 20 strictly increasing results, one for each choice of 3 different numbers from 6.

  15 is C(6, 2), the count of increasing pairs when a die is rolled twice, out of 36. A third roll changes both the count and the total.

• (2)
  1/18 — 1/18 = 12/216, which would mean 12 favourable outcomes.

  That is too few: the increasing results that start with 1 or 2 already number 10 + 6 = 16. Adding those starting with 3 and 4 gives 20, so the probability is 20/216 = 5/54.

• (3)
  13/216 — 13/216 has the right total, 216, but 13 favourable outcomes instead of 20.

  The 120 outcomes with three different numbers come in groups of 3! = 6 orders of the same numbers, and exactly one order in each group is increasing: 120 ÷ 6 = 20.

When outcomes are equally likely, probability = favourable outcomes ÷ total outcomes. Rolling a die n times gives 6ⁿ outcomes.

A strictly increasing sequence is decided by which numbers appear: once 3 different numbers are chosen, only one order is increasing. So the count is C(6, 3) = 20.

Allowing ties changes the count. Sequences in which each roll is at least the previous one number C(8, 3) = 56, choosing 3 numbers from 6 with repetition allowed.

By the same reasoning, strictly decreasing results also number 20.

Key facts

• Three rolls of a die have 6³ = 216 equally likely outcomes.
• Outcomes with three different numbers: 6 × 5 × 4 = 120.
• Each set of three different numbers occurs in 3! = 6 orders, exactly one increasing: 120 ÷ 6 = 20.
• P(each roll larger than the previous one) = 20/216 = 5/54.
• If each roll need only be at least the previous one, there are C(8, 3) = 56 outcomes, probability 7/27.

Common traps

• Counting arrangements. 6 × 5 × 4 = 120 counts every order of three different numbers; only 1 order in 6 is increasing.
• Allowing ties. "Larger than the previous number" excludes equal rolls; "at least" would give 56 outcomes.
• Multiplying step chances. P(second > first) = 15/36, but the chance that the third beats the second depends on the second, so (15/36)² does not give the answer.

• practice — not a real PYQ

  A die is rolled twice. What is the probability that the second number is larger than the first?

  • (a)5/12
  • (b)1/2
  • (c)1/6
  • (d)7/12
  Answer(1) — Of 36 outcomes, C(6, 2) = 15 have the second number larger: 15/36 = 5/12.

  Option (2) ignores the 6 ties; option (3) is the probability that both numbers are equal; option (4) counts the second number being larger or equal, 21/36.

• practice — not a real PYQ

  A die is rolled three times. What is the probability that all three numbers are different?

  • (a)5/9
  • (b)5/54
  • (c)1/2
  • (d)4/9
  Answer(1) — Outcomes with three different numbers: 6 × 5 × 4 = 120, so P = 120/216 = 5/9.

  Option (2) counts only the increasing order; option (3) would need 108 such outcomes; option (4) is the probability that at least two numbers are the same.