Probability 26

Any asnwer for reference?

2009-07-15 10:02:40 補充：
Number of ways 8 balls are put in 1 box without a box empty (P1) = 1 (all balls in the box)
Number of ways 8 balls are put in 2 boxes without a box empty (P2) =
Number of ways 8 balls are put in 2 boxes – (2C1)* number of ways 1 box is empty
= 28 – 2*1 = 254
Number of ways 8 balls are put in 3 boxes without a box empty (P3) = 38 – (3C1)*(1E2N) – (3C2)(2E1N) where E = Empty; N = Non-empty
= 38 – 3(254) – 3(1) = 5796
Number of ways 8 balls are put in 4 boxes without a box empty (P4) = 48 – (4C1)*(1E3N) – (4C2)*(2E2N) – (4C3)*(3E1N)
= 48 – 4*5796 – 6*254 – 4*1 = 40824
Number of ways 8 balls are put in 5 boxes without a box empty (P5)= 58 – (5C1)*(1E4N) – (5C2)*(2E3N) – (5C3)*(3E2N) – (5C4)*(4E1N)
= 58 – 5*40824 – 10*5796 – 10*254 – 5
= 126000
Number of ways 8 balls are put in 6 boxes without a box empty (P6) = 68 – (6C1)*(1E5N) – (6C2)*(2E4N) – (6C3)*(3E3N) – (6C4)*(4E2N) – (6C5)*(5E1N)
= 68 – 6*126000 – 15*40824 – 20*5796 – 15*254 – 6*1
= 191520
Probability of 6 empty boxes = 0
Probability of 5 empty boxes = (6C1)(P1)/68 = 6*1/1679616 = 0.0000036
Probability of 4 empty boxes = (6C2)(P2)/68 = 15*254/1679616 = 0.0022684
Probability of 3 empty boxes = (6C3)(P3)/68 = 20*5796/1679616 = 0.0690158
Probability of 2 empty boxes = (6C4)(P4)/68 = 15*40824/1679616 = 0.3645833
Probability of 1 empty boxes = (6C5)(P5)/68 = 6*126000/1679616 = 0.4501029
Probability of 0 empty boxes = (6C6)(P6)/68 = 1*191520/1679616 = 0.1140261

It is easy.

First, we should better call that the probability mass function of X.

f_X_(x) =

{ 191,520/1,679,616　for x = 0

756,000/1,679,616　for x = 1

612,360/1,679,616　for x = 2

115,920/1,679,616　for x = 3

3,810/1,679,616　for x = 4

6/1,679,616　for x = 5

0　for x = 6}

請原諒我暫時還是小學級,希望不要介意我不能用圖示..我已努力投票和評價加緊分

2009-07-14 19:10:48 補充：
你做這題從什麼concept出發?