M1 probability

philip ken

2012-06-15 3:37 am

two black balls and thirteen white balls are randomly put into five identical boxes,so each box contains three balls

a)find probability that two black balls are put in different boxes

b) A,B,C plat a game:
one box is randomly selected without replacement.A,B and C each in turn draws one ball randomly from the box without replacement.game stops when a black ball is drawn.person draws black ball wins game.
if no black ball has been drawn,the above process is repeated until a black ball is drawn.

i)let Pn be probability that exactly n balls are drawn in game.show Pn=1/7-n/105

II) calculate expected no if balls drawn in game

III)find prob.that A wins

回答 (1)

用戶2053

2012-06-15 4:35 am

✔ 最佳答案

a)Choose 2 boxes among 5 , 5C2 ways , then
3C1 ways for position in the box of each black ball.P(two black balls are put in different boxes)
= 5C2 * (3C1)² / 15C2
= 6/7
bi)Consider there are 15 - n possible positions for the 2nd black ball , Pn
= (15 - n) / 15C2
= (15 - n) / 105
= 1/7 - n/105
bii)Expected no. of balls drawn in game
= 1(1/7 - 1/105) + 2(1/7 - 2/105) + 3(1/7 - 3/105) + ... + 15(1/7 - 15/105)
= (1/7) (1+2+3+...+15) + (1/105) (1²+2²+3²+...+15²)
= (1/7) (1+15)15/2 + (1/105) (15*16*(2*15+1)/6)
= 120/7 + 248/21
= 608/21
biii)P(A win)
= P1 + P4 + P7 + P10 + P13
= (1/7 - 1/105) + (1/7 - 4/105) + (1/7 - 7/105) + (1/7 - 10/105) + (1/7 - 13/105)
= 5/7 - (1+4+7+10+13)/105
= 5/7 - 1/3
= 8/21

2012-06-15 01:54:47 補充：
Corr:

bii)

Expected no. of balls drawn in game
= 1(1/7 - 1/105) + 2(1/7 - 2/105) + 3(1/7 - 3/105) + ... + 15(1/7 - 15/105)
= (1/7) (1+2+3+...+15) - (1/105) (1²+2²+3²+...+15²)
= (1/7) (1+15)15/2 - (1/105) (15*16*(2*15+1)/6)
= 120/7 - 248/21
= 112/21
= 5 + 1/3

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