probability

2011-03-16 9:06 pm

There are 3 bags X,Y and Z. Bag X contains 1 red card and 3 green cards, bag Y contains 8 red cards and 4 green cards, and bag Z contains 2 red cards and 4 green cards. A card is drawn at ramdom from each of the 3 bags. It is known that 2 red cards and 1 green card care drawn. If a card is now drawn at random from bag Z, find the probability that it is red.

更新1:

Ans = 3/11, but do not know why?

更新2:

What is 56/88 What is 48+8+32 how do we get 7/11 and 4/11 ?

更新3:

how to get (1/5)(7/11) + (2/5)(4/11) ?

回答 (1)

myisland8132

2011-03-17 1:29 am

✔ 最佳答案

P(2 red cards and 1 green card)

= P(X green) + P(Y green) + P(z green)

= (3/4)(8/12)(2/6) + (1/4)(4/12)(2/6) + (1/4)(8/12)(4/6)

= 11/36

56/88

48+8+32

P(the probability that the second card drawn from Z is red)

= P(second card is red | first card of Z is red) + P(second card is red | first card of Z is green)

= (1/5)(7/11) + (2/5)(4/11)

= 15/55

= 3/11

2011-03-18 13:38:19 補充：
Please omit
56/88

48+8+32

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