IPA IP 2024 Question 83
A question IPA Japan published in 令和6年度 ITパスポート試験 公開問題 問83, translated into English. ITPEC is based on the same exams.
Same question on ITPEC: Apr 2025 Q4
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
This question comes from the past questions IPA Japan publishes, translated into English by ITPEC Practice. ITPEC Practice is an independent study tool and is not affiliated with IPA or ITPEC. See the original paper (PDF) or Report an issue.