推 Eureka7:第二題跟您見解有點不同，K=1時應該是end of the first 03/19 08:30
→ Eureka7:如此推算end of the third我覺得應該是K=3啦 03/19 08:31
→ Eureka7:其他題想法都差不多,但其實他每題都規定要用P表示 03/19 08:32
→ Eureka7:最討厭的就是這了 03/19 08:34
推 colan1206:非常感謝大大的答案^^" 03/19 10:01
※ 引述《colan1206 (嵐)》之銘言：
: 今年清大資應的機率論考題
: 困擾我很久>"<
: 希望有大大能幫忙解題
: 非常感激""
: 1.The probability that any paticular hard disk will fail during its Kth
: year of use is given by the probability mass function for K.
: P(K) = p * (1-p)^(K-1) ,K=1,2,3,...
: (Express the answer in terms of p.)
K~Geo(p)，p表硬碟壞掉的機率
: (1)Four hard disks are tested simultaneously.Determine the probability that
: none of the four disks fails during its first year of use.
P(K=1) = p ， 令 P1 = p
四個硬碟，第一年就壞掉的機率為P1
X~B(n=4.P1) = X~B(4.p)
P(X=0)= (4C0) * (p)^0 * (1-p)^4 = (1-p)^4
: (2)Four hard disks are tested simultaneously.Determine the probability that
: exactly two hard disks have faild by the end of the third year.
^^^^^^^^^^^^^^^^^^^^^^^^^
不知道是要指第四年壞掉
還是第三年壞掉
因為它的意思是第三年結束才壞掉
我個人覺得是 K=4
P(K=4) = p(1-p)^(3) ，令P2 = p(1-p)^(3)
X~B(n=4.P2)
P(X=2) = (4C2) * (P1)^2 * (1-P1)^2
: (3)Four hard disks are tested simultaneously.Determine the probability that
: exactly one hard disk fails during each of the first three years.
我覺得題意是指硬碟在"3年內"壞掉的機率
P(K<4) = P(K=1) + P(K=2) + P(K=3) 令P3 = p + p(1-p) + p(1-p)^2
= p + p(1-p) + p(1-p)^2
X~B(n=4.P3)
P(X=1) = (4C1) * (P3)^1 * (1-P3)^3
: (4)Four hard disks are tested simultaneously.Given that one hard disk has
: failed by the end of the first year,determine the probability that
: exactly two hard disks have failed by the end of the third year.
英文苦手
有點像是要考無記憶性，但是我看不太懂題意，答案看看就好...
四個中有一個在第一年就壞掉，有3個沒壞
剩下3個硬碟，第3年底(K=4)會壞掉的機率，表示它們還可以再使用2年的機率
P(K=4) = p(1-p)^2 令P4 = p(1-p)^2
X~B(n=3.P4)
P(X=2) = (3C2) * (P4)^2 * (1-P4)^1
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※ 編輯: royalwind 來自: 219.69.115.51 (03/19 00:32)