Two fair dice are thrown. Find the probability that at least one "6" occurs.
若果我以下咁做有錯嗎?
分了兩個cases
P(只抽到一個) + P(抽到兩個)
= (1/6 * 5/6) + (1/6 * 1/6)
Probability
2014-12-28 5:10 pm
回答 (2)
2014-12-29 12:01 am
✔ 最佳答案
Method1 :Pr(at least one "6")
= 1 - Pr(no "6")
= 1 - [1 - (1/6)]²
= 1 - (5/6)²
= 1 - (25/36)
= 11/36
Method 2 :
Pr(at least one "6")
= Pr(one "6") + Pr(two "6")
= Pr("6" and "not 6") + Pr("not 6" and "6") + Pr(two "6")
= (1/6) × [1 - (1/6)] + [1 - (1/6)] × (1/6) + (1/6)²
= (1/6) × (5/6) + (5/6) × (1/6) + (1/6)²
= (5/36) + (5/36) + (1/36)
= 11/36
2014-12-28 6:10 pm
P(只抽到一個) + P(抽到兩個)=(1/6*5/6+5/6*1/6)+(1/6)^2=11/36 It is because the order is important in this case.
收錄日期: 2021-04-15 17:43:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141228000051KK00010