Probability...stpm問題﹐救命的﹗希望各位可以幫忙﹗謝謝﹗

Harry7Kewell

an alarm system has 5 similar electronic components.The lifespan,in hours,of each electronic componenet which functions independently,is a rondom variable with the probability density function

         1/500^(e^-t/500)   ,t>0
  f(t)={ 0                  ,otherwise.

(i)Show that the expected lifespan of each electronic component in the alarm system is 500 hours.(解了)

(ii)Show that the probability that each electronic component in the alarm system will break down after operating for 500 hours is 0.632.(解了)

(iii)The alarm system will only function when not more than 2 of its electronic components are not working.Find the probability that the alrm system still work after operating 500 hours.(解了)

(iv)Find the probability that exactly 2 electronic components break down after operating for more than 500 hours,if the alarm system still works after operating for 500 hours.希望大家幫忙﹗

dunwan2tellu

我觉得(iv)的是conditional probability . 先找P(X=2) , X=number of break down component after 500 hour . 在除于(iii)所找到的probability 既可.

[ 本帖最后由 dunwan2tellu 于 5-11-2005 09:27 PM 编辑 ]

Harry7Kewell

是﹐你是對的。那是condition probability,所以(iv)的答案是﹕  

                    P{(X=2)n(X小于等于2)}
P(X=2/X小于等于2)= ---------------------
                         P(X小于等于2)

可是我就是找不到P{(X=2)n(X小于等于2)}~。~﹗﹗

其實這是stpm93年的問題﹗
供參考這是它的答案﹕(iii)0.2638
                    (iv)0.7446

dunwan2tellu

是 0.7547 吧? 只需要找 P(X=2)/P(X=<2) 就可以了。

P(X=2) = 5C2 (0.632)^2 (0.368)^3 = 0.19906
P(X=<2) = 0.26376

所以 P(X=2)/P(X=<2) =  0.7547

[ 本帖最后由 dunwan2tellu 于 5-11-2005 10:35 PM 编辑 ]

Harry7Kewell

...它的公式應該是
                                   P(AnB)
                          =>(A/B)=------
                                    P(B)

dunwan2tellu

你说得没错。但是 P{(X=2)n(X=<2)} = P(X=2) .... P(X=2)本身就是P(X=<2) 的 subset 罢了....

Harry7Kewell

原來如此﹐挺有道理。。。謝謝﹗