[數學]:機率問題

我自己本身讀英文 , 唔係好識中 文既terms

part a 答唔到你

首先你要define邊一個分佈~呢個係binomial (中文唔知咩)


b) 10 C2 ( 0.22)^2 (1-0.22)8

c)P(X >=1 ) = P(X = 0) + P(X =1 )
= (1-0.22) ^10 + 10 C1 (0.22 )( 1-0.22 )^9

i am not very sure about the chinese terms in maths, thus i cannot calculate the 標準差 for u

(a) 期望值 ( i guess is the no. of people expected to be 無法痊愈)
0.22 x 10
= 2.2
therefore, around 2 people 無法痊愈 within these 10 patients

(b) 兩名病人無法痊愈的機率
0.22^2 x (1 - 0.22)^8 x 45
= 0.29841 ( 5 sig fig)

(c) 至多一人無法痊愈的機率
0.22 x ( 1 - 0.22)^9 x 10 + ( 1 - 0.22)^10
= 0.31847 ( 5 sig fig)

2008-06-12 17:20:51 補充：
(a) 期望值 ( i guess is the no. of people expected to be 無法痊愈)
0.22 x 10
= 2.2
therefore, around 2 people 無法痊愈 within these 10 patients

(b) 兩名病人無法痊愈的機率
0.22^2 x (1 - 0.22)^8 x 45
= 0.29841 ( 5 sig fig)

(c) 至多一人無法痊愈的機率
0.22 x ( 1 - 0.22)^9 x 10 + ( 1 - 0.22)^10
= 0.31847 ( 5 sig fig)

i am not very sure about the chinese terms in maths, thus i cannot calculate the 標準差 for u

(a) 期望值 ( i guess is the no. of people expected to be 無法痊愈)
0.22 x 10
= 2.2
therefore, around 2 people 無法痊愈 within these 10 patients

(b) 兩名病人無法痊愈的機率
0.22^2 x (1 - 0.22)^8 x 45
= 0.29841 ( 5 sig fig)

(c) 至多一人無法痊愈的機率
0.22 x ( 1 - 0.22)^9 x 10 + ( 1 - 0.22)^10
= 0.31847 ( 5 sig fig)

2008-06-12 17:23:46 補充：
(a) 期望值 ( i guess is the no. of people expected to be 無法痊愈)
0.22 x 10
= 2.2
therefore, around 2 people 無法痊愈 within these 10 patients

(b) 兩名病人無法痊愈的機率
0.22^2 x (1 - 0.22)^8 x 45
= 0.29841 ( 5 sig fig)

(c) 至多一人無法痊愈的機率
0.22 x ( 1 - 0.22)^9 x 10 + ( 1 - 0.22)^10
= 0.31847 ( 5 sig fig)