i****b 发帖数: 52 | 1 When A predicts raining, the chance of raining is 60%. When B predicts
raining, the chance of raining is also 60%. If A and B both predict to rain
(assuming they are independent), what is the chance of raining? |
s*********y 发帖数: 284 | 2 缺条件...
or you can make some assumption
rain
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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i****b 发帖数: 52 | 3 nope. i asked whether the conditions are sufficient after 3 seconds and was
told yes.... |
h***s 发帖数: 2499 | 4 60%
rain
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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l******o 发帖数: 28 | 5 我猜是0.72。假设了无条件下下雨概率是0.5
不确定是否正确 等待牛人给出详细解答 |
i****b 发帖数: 52 | 6 为啥是60%?。。。。 我说我只能是确定0.6 ~ 0.84之间。。 。 然后面试官让我继
续做。。。 到底也没做出来。。。。。
【在 h***s 的大作中提到】 : 60% : : rain
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x***i 发帖数: 106 | 7 Stevens Capital Management?
我想假如没有下雨,就是both A and B are wrong,
If they are independent, P(A and B wrong) = P(A wrong)P(B wrong) = 16%
So P(rain) = 84%
我记得没有给出independent的条件,所以如果A和B是完全正相关的话,会是60%。所以
最后结果应该是一个Range。 |
i****b 发帖数: 52 | 8 是这个公司。。。原来你也被问过阿哈。。 我觉得不能假设他们independent, 因为A
错了,说明不下雨,那么B一定错。所以0.84的上线是错的。 0.6的下限是对的。
【在 x***i 的大作中提到】 : Stevens Capital Management? : 我想假如没有下雨,就是both A and B are wrong, : If they are independent, P(A and B wrong) = P(A wrong)P(B wrong) = 16% : So P(rain) = 84% : 我记得没有给出independent的条件,所以如果A和B是完全正相关的话,会是60%。所以 : 最后结果应该是一个Range。
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x***i 发帖数: 106 | 9 恩,有道理。那么根据类似于Naive Bayes的原理,应该是这样的:
为了简洁,考虑条件概率下的Sample space:
Sample space given both predict "rain" (2 samples):
1. A correct and B correct
2. A incorrect and B incorrect
P(A correct and B correct | prior)= 0.6
P(A incorrect and B incorrect | prior) = 0.14
P(rain | both predict "rain") = P(A correct and B correct in the "codition"
sample space) = Bayes rule like Naive Bayes... = 0.81
我记得他提到过答案比0.84稍微小。
A
【在 i****b 的大作中提到】 : 是这个公司。。。原来你也被问过阿哈。。 我觉得不能假设他们independent, 因为A : 错了,说明不下雨,那么B一定错。所以0.84的上线是错的。 0.6的下限是对的。
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c**********6 发帖数: 18 | 10 为什么P(A incorrect and B incorrect | prior) = 0.14 ?难道不是 0.4?
"
【在 x***i 的大作中提到】 : 恩,有道理。那么根据类似于Naive Bayes的原理,应该是这样的: : 为了简洁,考虑条件概率下的Sample space: : Sample space given both predict "rain" (2 samples): : 1. A correct and B correct : 2. A incorrect and B incorrect : P(A correct and B correct | prior)= 0.6 : P(A incorrect and B incorrect | prior) = 0.14 : P(rain | both predict "rain") = P(A correct and B correct in the "codition" : sample space) = Bayes rule like Naive Bayes... = 0.81 : 我记得他提到过答案比0.84稍微小。
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r*****d 发帖数: 346 | 11 Since you mentioned independence, I will go for 84%. It is not because 84%
is The answer, but it is sufficient to make a case.
rain
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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q*********i 发帖数: 21 | 12 有没有可能你听错题了,或者我理解错了。定义3个events, X=A predicts rain, Y=B
predicts rain, Z=actually rain。根据你的描述,已知条件是:P(Z|X)=0.6, P(Z|Y)
=0.6, P(XY)=P(X)P(Y),问题是:P(Z|XY)=?
这个答案从0到1都有可能,例子也不难构造。只要P(X),P(Y)足够小,P(Z|XY)想得多
少都可以。
rain
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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n******t 发帖数: 4406 | 13 我晕,金融界现在都这个level。
was
【在 i****b 的大作中提到】 : nope. i asked whether the conditions are sufficient after 3 seconds and was : told yes....
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N******K 发帖数: 10202 | 14 If A and B are two suckers, then nobody knows the chance of raining, could
be 0-1
If A and B are two gods, then 60% for sure
generally 60% +/- delta%
delta depends on the ability of the predictor
rain
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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l******n 发帖数: 1250 | 15 我想这是一个boost的典型问题,multipredictor,how to sum up.其实值肯定大于0.6
, 但具体多少,很难确定 |
r*********n 发帖数: 4553 | 16 我觉得这个答案比较靠谱
看了下上面的答案,我觉得A,B对下雨的prediction和真正下雨的概率是没有关系的
因为A,B是独立的,所以A作为B的side information是没有价值的(反之亦然),所以
最后A,B的prediction都没有变化,还都是60% within certain confidence interval
could
【在 N******K 的大作中提到】 : If A and B are two suckers, then nobody knows the chance of raining, could : be 0-1 : If A and B are two gods, then 60% for sure : generally 60% +/- delta% : delta depends on the ability of the predictor : : rain
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T*****u 发帖数: 7103 | 17 60%吧,confidence interval变窄了而已;把a和b想像成一个coin摔两遍好了。 |
k******n 发帖数: 12 | 18
could
我认为是0.8, 原因如下:
当A和B都说下雨就真的下雨的概率均匀分布在0.6到1.0之间,均匀分布是因为他们相互
独立;在这种情况下,中间结果等于最终结果,即最后结果是(0.6+1.0)/2=0.8
可以看看A的概率和B的概率重叠的情况:
P(rain | A predicts rain)=0.6, 这个0.6可以叫0.6A
P(rain | B predicts rain)=0.6, 这个0.6可以叫0.6B
0.6A和0.6B重叠是x的情况下,得到P(rain | A and B both predict rain)=0.6+0.6-x
x=0.6时,0.6A和0.6B在重叠最多,得到P(rain | A and B both predict rain)=0.6
x=0.2时,0.6A和0.6B在重叠最少,得到P(rain | A and B both predict rain)=1.0
x在0.2到0.6之间均匀分布,x取0.2到0.6间任意值的概率是1/0.4,所以最后结果是
把(1.2-x)*1/0.4在0.2到0.6之间对x积分,最终得到0.8
【在 N******K 的大作中提到】 : If A and B are two suckers, then nobody knows the chance of raining, could : be 0-1 : If A and B are two gods, then 60% for sure : generally 60% +/- delta% : delta depends on the ability of the predictor : : rain
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K*V 发帖数: 192 | 19 Most of u are over thinking. I thought the answer is 0.6*0.6=0.36.
When A predicts raining, the chance of raining is 60%. When B predicts
raining, the chan........
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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l****w 发帖数: 583 | 20 我也认为是60%。
其实用土办法想想也很直观:A和B预测下雨的能力是一样的(就是6成的命中率)。他
们都一同预测下雨并不影响各自及最终的命中率。下雨概率和预测下雨的命中概率在这
个题目里好像整得挺晕的。
看了一下,回答60%的都是我喜欢的ID。哈哈。
interval
【在 r*********n 的大作中提到】 : 我觉得这个答案比较靠谱 : 看了下上面的答案,我觉得A,B对下雨的prediction和真正下雨的概率是没有关系的 : 因为A,B是独立的,所以A作为B的side information是没有价值的(反之亦然),所以 : 最后A,B的prediction都没有变化,还都是60% within certain confidence interval : : could
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T*****u 发帖数: 7103 | 21 仁兄,如果问predict不下雨的概率捏?0.12? 那1-0.36-0.12=0.52 的概率属于啥?
predictor不工作概率?
【在 K*V 的大作中提到】 : Most of u are over thinking. I thought the answer is 0.6*0.6=0.36. : : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chan........
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c**********s 发帖数: 295 | 22 I seconde this.
B
Y)
【在 q*********i 的大作中提到】 : 有没有可能你听错题了,或者我理解错了。定义3个events, X=A predicts rain, Y=B : predicts rain, Z=actually rain。根据你的描述,已知条件是:P(Z|X)=0.6, P(Z|Y) : =0.6, P(XY)=P(X)P(Y),问题是:P(Z|XY)=? : 这个答案从0到1都有可能,例子也不难构造。只要P(X),P(Y)足够小,P(Z|XY)想得多 : 少都可以。 : : rain
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l***y 发帖数: 1166 | 23 P(AnB)=1-P(A'nB')
n means and
' means not |
K*V 发帖数: 192 | 24 P(AUB) = 1-(A'nB')=1-(1-0.6)*(1-0.6) = 0.86
【在 l***y 的大作中提到】 : P(AnB)=1-P(A'nB') : n means and : ' means not
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k*****h 发帖数: 12 | 25 认为答案是60%的同学,假如题目条件稍微改一下,改成A预测下雨,实际下雨的概率是
60%,B预测下雨,实际下雨的概率是80%,A B的预测互相独立,那么答案是?
【在 l****w 的大作中提到】 : 我也认为是60%。 : 其实用土办法想想也很直观:A和B预测下雨的能力是一样的(就是6成的命中率)。他 : 们都一同预测下雨并不影响各自及最终的命中率。下雨概率和预测下雨的命中概率在这 : 个题目里好像整得挺晕的。 : 看了一下,回答60%的都是我喜欢的ID。哈哈。 : : interval
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k*****h 发帖数: 12 | 26 second这位仁兄。个人觉得应该是这样求解,求指正:
已知:
P(Z|X) = 0.6
P(Z|Y) = 0.6
P(XY) = P(X)P(Y)
求P(Z|XY)
概率公式:
P(Z) = P(Z|X)P(X) (1)
P(Z) = P(Z|Y)P(Y) (2)
P(Z) = P(Z|XY)P(XY) = P(Z|XY)P(X)P(Y) (3)
(1)*(2)/(3)得
P(Z|XY) = P(Z|X)*P(Z|Y)/P(Z)=0.6*0.6/P(Z)
P(Z)未知,所以无法求解。
B
Y)
【在 q*********i 的大作中提到】 : 有没有可能你听错题了,或者我理解错了。定义3个events, X=A predicts rain, Y=B : predicts rain, Z=actually rain。根据你的描述,已知条件是:P(Z|X)=0.6, P(Z|Y) : =0.6, P(XY)=P(X)P(Y),问题是:P(Z|XY)=? : 这个答案从0到1都有可能,例子也不难构造。只要P(X),P(Y)足够小,P(Z|XY)想得多 : 少都可以。 : : rain
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k*****h 发帖数: 12 | 27 这位同学,你对P(A wrong)的理解不对
P(A wrong)有两种情况,一种是预报下雨却没下,一种是预报没下却下了。你只考虑了
一种情况。
【在 x***i 的大作中提到】 : Stevens Capital Management? : 我想假如没有下雨,就是both A and B are wrong, : If they are independent, P(A and B wrong) = P(A wrong)P(B wrong) = 16% : So P(rain) = 84% : 我记得没有给出independent的条件,所以如果A和B是完全正相关的话,会是60%。所以 : 最后结果应该是一个Range。
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z****e 发帖数: 19 | 28 过程看起来没问题,不过最后结果假设p(z) = 0.1, p(z|xy) = 3.6?
【在 k*****h 的大作中提到】 : second这位仁兄。个人觉得应该是这样求解,求指正: : 已知: : P(Z|X) = 0.6 : P(Z|Y) = 0.6 : P(XY) = P(X)P(Y) : 求P(Z|XY) : 概率公式: : P(Z) = P(Z|X)P(X) (1) : P(Z) = P(Z|Y)P(Y) (2) : P(Z) = P(Z|XY)P(XY) = P(Z|XY)P(X)P(Y) (3)
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a*******e 发帖数: 253 | 29 难道不是p(z,x)=p(z/x)p(x),还是我记错了?
【在 k*****h 的大作中提到】 : second这位仁兄。个人觉得应该是这样求解,求指正: : 已知: : P(Z|X) = 0.6 : P(Z|Y) = 0.6 : P(XY) = P(X)P(Y) : 求P(Z|XY) : 概率公式: : P(Z) = P(Z|X)P(X) (1) : P(Z) = P(Z|Y)P(Y) (2) : P(Z) = P(Z|XY)P(XY) = P(Z|XY)P(X)P(Y) (3)
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D********e 发帖数: 8 | 30 P(not rain) = P(A and B wrong) = P(A wrong)P(B wrong) = 16%
P(rain) = P(A and B right) = P(A right)P(B right) = 36%
The answer = 36/(36+16) = 36/52
Any comment? |
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D********e 发帖数: 8 | 31 This problem is about given a and b give the same prediction, what is the
chance that they are both right to wrong.
The answer is clearly 50% if both prediction accuracy is 50%. So any
solution does not agree with special case won't be right.
If a 's accuracy is 50%, then the final answer should be the same as the b '
s prediction accuracy. Again any solution does not agree with this case won
't be right.
At least I. Am convinced that I am right.
【在 D********e 的大作中提到】 : P(not rain) = P(A and B wrong) = P(A wrong)P(B wrong) = 16% : P(rain) = P(A and B right) = P(A right)P(B right) = 36% : The answer = 36/(36+16) = 36/52 : Any comment?
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H**********l 发帖数: 12 | 32 P(raining|A predicts raining)*P(raining|B predicts raining)/P(raining)
rain
【在 i****b 的大作中提到】 : When A predicts raining, the chance of raining is 60%. When B predicts : raining, the chance of raining is also 60%. If A and B both predict to rain : (assuming they are independent), what is the chance of raining?
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