Question

求解一道概率题~！

Andrei Markov, a Russian spy, has three Americans – Joe, John, and Jim – under surveillance. Each man owns one and only one car. Markov knows that the three cars are painted blue, black, and brown respectively, but does not know which man owns which car. After considerable snooping, he estimates the probabilities of each man owning each of the cars as follows:

Car
Blue Black Brown
Joe .5 0 .5
John .25 .25 .5
Jim .25 .75 0

After Markov has determined three probabilities, Kolmogorov a trusted informer, tells him that Joe owns the brown car. With this information, what is the probability that Jim owns that black car?

Answer 1

Joe owns the brown car,也就是说其概率变成1，另外Joe拥有Blue和Black的概率都变成0，且John和Jim拥有brown car的概率也变成0。

另外很重要的一点是每一行每一列的和都应该是1，这个不能变。

Car
Blue Black Brown
Joe 0 0 1
John x 1-x 0
Jim 1-x . x 0
然后我们将关注点放在John身上，之前Markov估计他拥有Brown车的概率是0.5，Blue和Black相等，都是0.25.现在Brown变成0，那么Blue和Black就都是0.5，所以x=1-x=0.5
所以Jim拥有黑车的概率是0.5

Answer 2

由题意可知，白色车占全市车的1/5，
也就是说白色车出现的概率是1/5，
因为目击者看清楚该车颜色的概率是3/10
那么是白车的概率是：3/10×1/5=3/50

Answer 3

设A={白色车肇事}，B={目击者看到白色车肇事}
由贝耶式公式
P(A|B)=P(AB)/P(B)=P(A)P(B|A)/[P(A)P(B|A)+P(A')P(B|A')]
注：A'表示A的补集
其中P(A)=1/5,P(A')=4/5,P(B|A)=0.3,P(B|A')=0.7
代入得
P(A|B)=3/31

Answer 4

设事件A=车是白色，B=车被看成白色。则：非A=车是蓝色
非B=车被看成蓝色
由全概率公式：
P(B)=P(A)P(B|A)+P(非A)P(B|非A)
=0.2×0.3+0.8×0.7=0.62
P(A|B)=P(AB)/P(B)
=P(A)P(B|A)/P(B)
=0.2×0.3/0.62=3/31

Answer 5

我怎么认为分两种情况就行。一种在方形内可直接积分，另一种三角积分加上方形的