In a bag, there are 50 coins, in which 49 of them is normal, namely with one side head and the other side tail. And there is only one abnormal, with heads on both sides.
It is now drawn 6 coins randomly and tossed first five of them, it is known that the five coins tossed show heads. Find the probability that the sixth coin drawn is the abnormal coin.
The answer is 1/55.
Thanks in advance.
Difficult Probability
2013-03-25 12:40 am
回答 (4)
2013-03-25 6:49 am
✔ 最佳答案
--------------------------------------------------------------------------------------------------圖片參考：http://imgcld.yimg.com/8/n/HA05107138/o/20130324224915.jpg
2013-03-26 3:26 am
Hypercube, I want to know your Math background too.
Simon YAU
2013-03-26 09:41:31 補充：
I major in Math in U and plan to study MSc in Math in CU.
Simon
Simon YAU
2013-03-26 09:41:31 補充：
I major in Math in U and plan to study MSc in Math in CU.
Simon
2013-03-26 12:41 am
"What is your Math background?"
I want to know too.
2013-03-25 20:52:46 補充：
I'm just an amateur maths lover.
How about you?
I want to know too.
2013-03-25 20:52:46 補充：
I'm just an amateur maths lover.
How about you?
2013-03-25 7:23 am
Alternative method :
N : normal coin
A : abnormal coin
H : head
P(first 5 coins show H)
= P(5N and 5H) + P(4N1A and 5H)
= (49C5 / 50C5) x (1/2)^5 + (49C4 x 1C1 / 50C5) x (1/2)^4 x 1
= (9/320) + (1/160)
= 11/320
2013-03-24 23:24:14 補充：
P([first 5 coins shown H] and [6th coin is A])
= P([first 5 coins are 5N and 5H] and [6th coin is A]
= [(49C5 / 50C5) x (1/2)^5] x (1C1 / 45C1)
= (9/320) x (1/45)
= 1/1600
2013-03-24 23:24:48 補充：
The required probability
= P([6th coin is A] | [first 5 coins shown H])
= P([first 5 coins shown H] and [6th coin is A]) / P(first 5 coins show H)
= (1/1600) / (11/320)
= 1/55 ..... ans
N : normal coin
A : abnormal coin
H : head
P(first 5 coins show H)
= P(5N and 5H) + P(4N1A and 5H)
= (49C5 / 50C5) x (1/2)^5 + (49C4 x 1C1 / 50C5) x (1/2)^4 x 1
= (9/320) + (1/160)
= 11/320
2013-03-24 23:24:14 補充：
P([first 5 coins shown H] and [6th coin is A])
= P([first 5 coins are 5N and 5H] and [6th coin is A]
= [(49C5 / 50C5) x (1/2)^5] x (1C1 / 45C1)
= (9/320) x (1/45)
= 1/1600
2013-03-24 23:24:48 補充：
The required probability
= P([6th coin is A] | [first 5 coins shown H])
= P([first 5 coins shown H] and [6th coin is A]) / P(first 5 coins show H)
= (1/1600) / (11/320)
= 1/55 ..... ans
收錄日期: 2021-04-13 19:23:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130324000051KK00188