maths question

2012-09-07 5:28 am

.There are 4 urns w,x,y and z,where b,g and r stand for blue balls.green balls and red balls respectively.
urn w contains 1 blue ball, 1 green ball and 1 red ball.
urn x contains 1 green ball and 1 red ball.
urn y contains 2 blue balls and 1 green ball.
urn z contains 1 blue ball and 2 green balls

a)sam draws one ball randomly fron urn w and then put it into urn x.find the probabilities that

i)there are two red balls in urn x
ii)there is a red ball in urn w

b)after sam has put the ball drawn into urn x as described in (a).ling chooses one urn at random and then randomly draws one ball from the urn.find probabilities that
i)the ball drawn by sam and ling are both red,
ii)the balls drawn by sam and ling are of the same colour.

回答 (1)

50418129

2012-09-07 5:46 pm

✔ 最佳答案

(a)(i)

P(two red balls in urn x)
= P(draws red ball from urn w)
= 1/3

(a)(ii)

P(1 red ball in urn w)
= P(draws green ball or blue ball from urn w)
= 2/3

(b)(i)

P(sam and ling draw red balls)
= P(sam draws red ball from urn w) x
[P(ling draws red ball from urn w and ling choose urn w)
+ P(ling draws red ball from urn x and ling choose urn x)
+ P(ling draws red ball from urn y and ling choose urn y)
+ P(ling draws red ball from urn z and ling choose urn z)]
= 1/3 x [(0)(1/4) + (2/3)(1/4) + (0)(1/4) + (0)(1/4)]
= 1/18

(b)(ii)

P(sam and ling draw same colour balls)
= P(sam and ling draw red balls)
+ P(sam and ling draw green balls)
+ P(sam and ling draw blue balls)
= 1/3 x [(0)(1/4) + (2/3)(1/4) + (0)(1/4) + (0)(1/4)]
+ 1/3 x [(0)(1/4) + (2/3)(1/4) + (1/3)(1/4) + (2/3)(1/4)]
+ 1/3 x [(0)(1/4) + (1/3)(1/4) + (2/3)(1/4) + (1/3)(1/4)]
= 1/18 + 5/36 + 1/9
= 11/36

參考: knowledge

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