[機統] 關於期望值

看板Math作者sky428 (人)時間15年前 (2011/01/23 12:49)推噓0(0推 0噓 9→)

留言9則, 2人參與討論串1/2 (看更多)

Question: A fair die is rolled repeatedly. Let X be the number of rolls needed to obtaina 5 and Y the number of rolls needed to obtain a 6. Calculate E(X|Y = 2). 下面是他給的解答 Solution: X follows a geometric distribution with p = 1/6. Y = 2 implies the first roll is not a 6 and the second roll is a 6. This means a 5 is obtained for the first time on the first roll (probability = 20%) or a 5 is obtained for the first time on the third or later roll (probability = 80%). E(X|X>=3) = 1/p + 2 = 6+2 = 8, so E(X|Y=2) = .2(1) + .8(8) = 6.6 我機率沒學好，英文也很糟，所以看不懂>"< 我的疑問是... 0.2是怎麼來的??他所代表的意思是P(X=1|Y=2)?? (是因為第一次擲到的可能只剩下點數1~5，所以是1/5?) 為什麼不是(1/5)*(1/6).......我知道我哪裡想錯了 P(X=1|Y=2)=(1/6)*(1/6)/(5/6)*(1/6)=1/5 E(X|X>=3) = 1/p + 2 = 6+2 = 8 ﹋﹋﹋﹌﹌﹌﹋﹋﹋﹋這地方不懂 E(X|Y=2) = .2(1) + .8(8) = 6.6 拜託解答一下，謝謝!!!! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.227.138.102

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