Probability
2010-05-29 9:04 pm
A bag with 3 yellow sweets, 5 green sweets and 4 blue sweets (total 12). To draw one at a time, what is the probability of getting 1st blue, 2nd blue 3rd any color and 4th yellow?
回答 (2)
2010-05-29 10:18 pm
✔ 最佳答案
P(E)= P(blue) * P(blue) * P(any color) * P(yellow)
= P(blue) * P(blue) * P(yellow) * P(yellow)
+ P(blue) * P(blue) * P(not yellow) * P(yellow)
= 4/12 * 3/11 * 3/10 * 2/9 + 4/12 * 3/11 * 7/10 * 3/9
= 324/11880
= 27/990
2010-05-29 14:19:26 補充：
= 3/110
2010-05-30 12:58 am
P(getting 1st blue)
= 4/12
=1/3
P(getting 2nd blue)
= 3/11
P(getting 3rd any color)
= 1
P(getting 4th yellow)
= 2/9
P(required result)
=(1/3)*(3/11)*(1)*(2/9)
=2/99
= 4/12
=1/3
P(getting 2nd blue)
= 3/11
P(getting 3rd any color)
= 1
P(getting 4th yellow)
= 2/9
P(required result)
=(1/3)*(3/11)*(1)*(2/9)
=2/99
參考: me
收錄日期: 2021-04-21 22:11:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100529000051KK00614