Probability
2012-12-17 6:20 pm
There are 4 black balls, 2 white balls and 1 blue ball in a bag. One ball is drawn at a time at random from the bag without replacement. Find the probability that it takes at least 3 draws to get a black ball.
回答 (3)
2012-12-18 11:56 pm
✔ 最佳答案
P( takes at least 3 draws)=P(takes more than or equal to 3 draws)
=P(1- 1 draw -2 draws)
=1- 4/7 -(3/7)*(4/6)
=1-4/7 -2/7
=1/7 //
2012-12-17 8:09 pm
Sol
A：第一次不是黑球
B：第二次不是黑球
C：第三次不是黑球
D：第三是黑球
E：第四次是黑球
P=P(ABCE)+P(ABD)
P(ABCE)
=P(E|ABC)*P(ABC)
=P(E|ABC)*P(C|AB)*P(AB)
=P(E|ABC)*P(C|AB)*P(B|A)*P(A)
=1*(1/5)*(2/6)*(3/7)
P(ABD)
=P(D|AB)*P(AB)
=P(D|AB)*P(B|A)*P(A)
=(4/5)*(2/6)*(3/7)
P=1*(2/6)*(3/7)=1/7
A：第一次不是黑球
B：第二次不是黑球
C：第三次不是黑球
D：第三是黑球
E：第四次是黑球
P=P(ABCE)+P(ABD)
P(ABCE)
=P(E|ABC)*P(ABC)
=P(E|ABC)*P(C|AB)*P(AB)
=P(E|ABC)*P(C|AB)*P(B|A)*P(A)
=1*(1/5)*(2/6)*(3/7)
P(ABD)
=P(D|AB)*P(AB)
=P(D|AB)*P(B|A)*P(A)
=(4/5)*(2/6)*(3/7)
P=1*(2/6)*(3/7)=1/7
2012-12-17 6:41 pm
For it to take at least 3 draws to get a black ball, the first 2 must be non-black ball with probability:
3/7 x 2/6 = 1/7
(Since in the first draw, there are 7 balls in which there are 3 not black and in the 2nd draw, there are 6 balls in which there are 2 not black)
3/7 x 2/6 = 1/7
(Since in the first draw, there are 7 balls in which there are 3 not black and in the 2nd draw, there are 6 balls in which there are 2 not black)
參考: 原創答案
收錄日期: 2021-04-24 09:52:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121217000051KK00062