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Suppose we know there is a ball X in the pocket, but do not know the color of the ball (we know its color is white or black). One puts another ball Y with white color into the pocket, then takes out a ball randomly. If the ball taken out is white, What is the probability that X is white?

1 Answers 1

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This is a straightforward problem using Baye's method:

Let $A$ be the event that the ball taken out is white and $B$ be the event that $X$ is white.

You wish to compute $P(B\mid A)$. This isn't straightforward to do; but $P(A\mid B)$ is easy to compute. It would thus be desirable to switch the roles of $A$ and $B$. This is effected as follows:

$\tag{1} \eqalign{ P(B\mid A)&={P(A\cap B)\over P(A)} ={P(A\mid B)P(B)\over P(A)}.\cr } $ Now to find $P(A)$, we condition on what color $X$ is: $ P(A)= P(A\cap B)+P(A\cap B^C)=P(A\mid B)P(B)+P(A\mid B^C)P(B^C) . $ Substituting the above into $(1)$, we obtain: $\tag{2} \eqalign{ P(B\mid A) &={P(A\mid B)P(B)\over P(A\mid B)P(B)+P(A\mid B^C)P(B^C) }. } $

Now:

  • $P(A\mid B)$ is the probability that the ball taken out is white given that $X$ is white; so, $P(A\mid B)=1$.
  • $P(A\mid B^C)$ is the probability that the ball taken out is white given that $X$ is not white; so, $P(A\mid B^c)=1/2$.
  • Presumably $P(B)=1/2$. Then $P(B^c)=1/2$.

Now it's a matter of substituting this information into the right hand side of $(2)$.