問83
頻出度★★☆
1から6までの六つの目をもつサイコロを3回投げたとき，1回も1の目が出ない確率は幾らか。
令和6年度 ITパスポート試験 問83
テクノロジ系基礎理論 › 応用数学
This question's explanation and your answers so far.
解説
問 83
Incorrectア
1/216 is (1/6)^3, the probability that a 1 comes up on all three throws. The question asks for the probability that a 1 never comes up at all, which is the opposite event, so this is wrong.
Incorrectイ
5/72 (= 15/216) corresponds to the probability that a 1 comes up exactly twice in three rolls, 3 × (1/6)² × (5/6), not the probability that a 1 never comes up. The required probability is (5/6)³ = 125/216, so this is wrong.
Incorrectウ
91/216 equals 1 − 125/216, which is the probability that a 1 comes up at least once in the three throws (the complementary event). This is the opposite of the required probability that a 1 never comes up, so it is wrong.
- エ
正解 When a die is thrown once, the probability of “not getting a 1” is 5/6, because 5 of the 6 possible outcomes are not a 1. The three trials are independent of one another, so the probability that no 1 appears in any of the three throws is found by multiplying them: (5/6)×(5/6)×(5/6) = (5/6)^3 = 125/216. Therefore the correct answer is choice エ, 125/216.
ここがポイント
- The probability of independent trials occurring one after another is found by multiplying the probability of each trial (the multiplication rule). Here, multiplying (5/6) three times gives (5/6)^3 = 125/216.
- “Never appears” and “appears at least once” are complementary events (their two probabilities sum to 1). Be careful not to confuse 125/216 with 91/216.
用語