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I was given the following question (here I give only the setting):

A box has $10$ red balls and $5$ black balls. A ball is selected from the box. If the ball is red, it is returned to the box. If the ball is black, it is returned to the box along with two additional black balls. After this, another ball is drawn.

To solve one of parts in the question I wish to calculate $P(B^{c}|A^{c})$ where $A$ as the event that the first ball drawn is Red, $B$ as the event that second ball drawn is Red and given that I calculated $$P(A)=\frac{2}{3},P(B)=\frac{98}{153},P(B|A^{c})=\frac{10}{17}$$

Given that $A^{c}$ occurs I have it that before the second draw there are $7$ Blcak balls and $10$ Red balls so the probability of drawing a black ball is $\frac{7}{17}=1-P(B|A^{c})$.

But if $$P(B^{c}|A^{c})=1-P(B|A^{c})$$ then $$P(B^{c}|A^{c})+P(B|A^{c})=1$$

But $A^{c}$ may not occur so I believe that the last equality is false.

What am I doing wrong here ?

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1 Answer 1

up vote 3 down vote accepted

You’re doing nothing wrong in the calculations; your error is in thinking that $$P(B^{c}\mid A^{c})+P(B\mid A^{c})=1\tag{1}$$ is contradicted by the fact that $A^c$ may not occur. $P(B^c\mid A^c)$ is the probability of $B^c$ given that $A^c$ occurs, and $P(B\mid A^c)$ is the probability of $B$ given that $A^c$ occurs. If you know that $A^c$ has occurred, then indeed it is true that either $B^c$ or $B$ must occur, which is exactly what $(1)$ says.

Now if you had somehow arrived at the equation $$P(B^c\text{ and }A^c)+P(B\text{ and }A^c)=1\;,\tag{2}$$

then you’d be right to be concerned, and for exactly the reason that you gave: $A^c$ may not occur. But $P(B\text{ and }A^c)$ is different from $P(B\mid A^c)$; in fact, $P(B\text{ and }A^c)=P(B\mid A^c)P(A^c)$.

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Thank you Brian! –  Belgi Nov 11 '12 at 19:07
    
@Belgi: You’re welcome! –  Brian M. Scott Nov 11 '12 at 19:10
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