Probability 難題

1, 應該係P(A)+P(B)-2*P(A intersect B)..
你可以畫個vann diagram...就會睇到...
如果係A or B or both...咁就係P(A)+P(B)-P(A intersect B)
但如果係not both...就要再減中間o個part...

2,
a, (990/1000)*(989/999)*(988/998)*(987/997)
b, (990/1000)*(989/999)*(988/998)*(10/997)

2007-09-10 00:07:59 補充：
第二題個答案其實好depend on with/without replacement...即係o個four parts 係拎左一個上黎之後放返落去...再拎第二個上黎 : with replacement一次過拎曬4個上黎 : without replacement....我個答案就係當without replacement...因為題目冇特別指明有冇replacement...而我個人覺得first four parts chosen for inspection多數都係without replacement...考試應該會清楚d...

2007-09-11 06:27:40 補充：
係venn diagram至啱

2007-09-11 13:58:27 補充：
hello, 第一題, 我唔明點解 P(A intersect B)要x 2? 即係減 2 次﹖冇錯..係要減兩次...因為你畫左venn diagram, 就會見到intersect ge 部份重疊左而你係要A or B but not both咁intersect部份即係唔要la...咁你咪要減兩次先可以un-count左not both的probability lor

2007-09-11 14:05:29 補充：
比個具體例子你A: 擇骰仔擇到2,4,6 B: 擇骰仔擇到1,2,3咁樣ge話P(A) = 1/2, P(B) = 1/2, P(A intersect B) = 1/6 (因為擇到2先會AB同時發生)咁我問你A or B but not both 係幾多你就可以用睇venn diagram ge方法...P(A) P(B)-2P(intersect) = 1/2＋1/2-2*1/6 = 2/3驗證：A or B but not both，即係擇到1,3,4,64/6 = 2/3... correct!!

1 Answer is P(A) + P(B) - P(A intersects B)

2 (a) 0.99^4 = 0.960596
(b) 0.99^3*0.01 = 0.0097