ITPEC IP April 2025 Question 4
Topic: Basic Theory
When a six (6)-sided dice with the numbers 1 to 6 is thrown three (3) times, what is the probability of not getting any 1?
Explanation
125/216 — each of the three independent throws must show one of the five faces 2 to 6.
- •One throw without a 1: 5 of 6 faces, probability 5 ÷ 6
- •Three independent throws: (5 ÷ 6)3 = 125 ÷ 216
Counted directly: of the 63 = 216 equally likely sequences, 53 = 125 use only the faces 2 to 6.
Answer (d)
Why not others:
- •(a) 1/216 — (1 ÷ 6)3, the probability of three 1s, the opposite extreme
- •(b) 5/72 — equals 15/216, the probability of exactly two 1s: 3 × (1 ÷ 6)2 × (5 ÷ 6)
- •(c) 91/216 — 1 − 125 ÷ 216, the complement: the probability of getting at least one 1
Key rule: For "none in n tries", raise the single-try probability of failure to the n-th power; "at least one" is 1 minus that.
Basic Theory · AI-written, may contain errors
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