m****r
发帖数: 141 | 1 This is an interview question of probability. The interview has been done.
In a room, there are 8 drawers in a desk, a man put a letter into one of
them with 50% randomly. After that, you enter the room, if drawers from 1 to
7 have been opened to show they are empty, what is probability of the 8th
drawer has a letter ?
My idea:
this is a conditional probability question and can be solved by Bayes
formula.
But, I got result 14/29 different from interviewer's answer.
Any help will be appreciated.
thanks | x*****t
发帖数: 353 | 2 P(d8=1|d1=0...d7=0) = ?
P(d8=1|d1=0,...d7=0) * P(d1=0,...d7=0) = P(d8=1,d1=0,...d7=0)=(1/2) * (1/8)=
1/16
P(d1=0,...d7=0) = P(d1=0,...d7=0,d8=0) + P(d1=0,...d7=0,d8=1)
= 1/2 + (1/2)*(1/8) = 9/16
So my answer is 1/9
to
【在 m****r 的大作中提到】
: This is an interview question of probability. The interview has been done.
: In a room, there are 8 drawers in a desk, a man put a letter into one of
: them with 50% randomly. After that, you enter the room, if drawers from 1 to
: 7 have been opened to show they are empty, what is probability of the 8th
: drawer has a letter ?
: My idea:
: this is a conditional probability question and can be solved by Bayes
: formula.
: But, I got result 14/29 different from interviewer's answer.
: Any help will be appreciated.
| c**********e
发帖数: 2007 | 3 correct.
)=
【在 x*****t 的大作中提到】
: P(d8=1|d1=0...d7=0) = ?
: P(d8=1|d1=0,...d7=0) * P(d1=0,...d7=0) = P(d8=1,d1=0,...d7=0)=(1/2) * (1/8)=
: 1/16
: P(d1=0,...d7=0) = P(d1=0,...d7=0,d8=0) + P(d1=0,...d7=0,d8=1)
: = 1/2 + (1/2)*(1/8) = 9/16
: So my answer is 1/9
:
: to
| a*****k
发帖数: 704 | 4 not quite understand "a man put a letter into one of
to | s******u
发帖数: 676 | 5 what is the 50% mean here?
to
【在 m****r 的大作中提到】
: This is an interview question of probability. The interview has been done.
: In a room, there are 8 drawers in a desk, a man put a letter into one of
: them with 50% randomly. After that, you enter the room, if drawers from 1 to
: 7 have been opened to show they are empty, what is probability of the 8th
: drawer has a letter ?
: My idea:
: this is a conditional probability question and can be solved by Bayes
: formula.
: But, I got result 14/29 different from interviewer's answer.
: Any help will be appreciated.
| c**********e
发帖数: 2007 | 6 With 50% probability, the ball is put in one of the 8 draws.
【在 a*****k 的大作中提到】
: not quite understand "a man put a letter into one of
: to
| r****t
发帖数: 10904 | 7 agreed
)=
【在 x*****t 的大作中提到】
: P(d8=1|d1=0...d7=0) = ?
: P(d8=1|d1=0,...d7=0) * P(d1=0,...d7=0) = P(d8=1,d1=0,...d7=0)=(1/2) * (1/8)=
: 1/16
: P(d1=0,...d7=0) = P(d1=0,...d7=0,d8=0) + P(d1=0,...d7=0,d8=1)
: = 1/2 + (1/2)*(1/8) = 9/16
: So my answer is 1/9
:
: to
| e*********r
发帖数: 80 | | x**********2
发帖数: 169 | | S******a
发帖数: 1426 | 10 楼上的都好聪明，我算了两遍才算到1/9,都忘光了 | k***t
发帖数: 33 | 11 Conditional probability gives 1/9, but suppose we conduct the experiment one
million times, and each time the opened 7 drawers are empty, what is the
expected number
of times that one finds the ball in the 8th drawer? Confused. | d**0
发帖数: 124 | 12 could you explain why P(d1=0,...d7=0) = 1/8? thx
)=
【在 x*****t 的大作中提到】
: P(d8=1|d1=0...d7=0) = ?
: P(d8=1|d1=0,...d7=0) * P(d1=0,...d7=0) = P(d8=1,d1=0,...d7=0)=(1/2) * (1/8)=
: 1/16
: P(d1=0,...d7=0) = P(d1=0,...d7=0,d8=0) + P(d1=0,...d7=0,d8=1)
: = 1/2 + (1/2)*(1/8) = 9/16
: So my answer is 1/9
:
: to
| n****e
发帖数: 629 | 13 假设有16个抽屉，立得1/9
to
【在 m****r 的大作中提到】
: This is an interview question of probability. The interview has been done.
: In a room, there are 8 drawers in a desk, a man put a letter into one of
: them with 50% randomly. After that, you enter the room, if drawers from 1 to
: 7 have been opened to show they are empty, what is probability of the 8th
: drawer has a letter ?
: My idea:
: this is a conditional probability question and can be solved by Bayes
: formula.
: But, I got result 14/29 different from interviewer's answer.
: Any help will be appreciated.
| d******w
发帖数: 102 | 14 这个真是神解法，膜拜大牛
【在 n****e 的大作中提到】
: 假设有16个抽屉，立得1/9
:
: to
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