# Conditional Probability?

I have a homework question that states that Bowl A has four red and two white chips and that Bowl B has three red and two white chips. A chip is drawn from random from bowl A and put into bowl B. After the chip is put into bowl B what is the probability that I draw a red chip from bowl B.

I have tried this: $$\frac{2}{6}\times\frac{3}{5}+\frac{4}{6}\times\frac{4}{5}$$

But got the answer wrong so I figure I'm on the right track but still did something the wrong way.

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From which bowl are you drawing the chip? – David Mitra Feb 2 '13 at 22:41
I am drawing from Bowl A – Brian Hauger Feb 2 '13 at 22:45
There is a $\color{maroon}{4\over6}$ chance that bowl $A$ winds up with three red chips and two white chips, and a $\color{darkgreen}{2\over6}$ chance that it winds up with four red chips and one white chip. The probability that you chose red is $\color{maroon}{4\over 6}\cdot {3\over5}+\color{darkgreen}{2\over6}\cdot{4\over5}$. – David Mitra Feb 2 '13 at 22:49
Oh wait I just got what you meant by your first comment. It is bowl B that I draw the final chip from. – Brian Hauger Feb 2 '13 at 22:53
Can you see where you went wrong now? (Bowl $B$ has a $4/6$ chance of winding up with four red chips and two white chips, and a $2/6$ chance of winding up with three red chips and three white chips.) – David Mitra Feb 2 '13 at 22:58

It seems you miscounted the total number of chips in bowl $B$ after the chip from bowl $A$ was put in.

After the chip taken from bowl $A$ is put into bowl $B$:

• The probability that bowl $B$ has four red chips and two white chips is $4/6$ (this is just the probability that a red chip was initially selected from bowl $A$).
• The probability that bowl $B$ has three red chips and three white chips is $2/6$ (this is just the probability that a white chip was initially selected from bowl $A$).

Let $R$ be the event that you chose a red chip from bowl $B$ after the chip from bowl $A$ was put in.

$P(R)$ can be found by conditioning on what type of chip was initially chosen from bowl $A$:

$\ \ \$Let $A_r$ be the event that the chip chosen from bowl $A$ and put into bowl $B$ was red.

$\ \ \$Let $A_w$ be the event that the chip chosen from bowl $A$ and put into bowl $B$ was white.

Then \eqalign{ P(R)&=P(R\cap A_r)+P(R\cap A_w)\cr &=P(A_r)P(R\,|\,A_r)+P(A_w)P(R\,|\,A_w)\cr &=\textstyle{4\over 6}\cdot {4\over6} +{2\over 6}\cdot {3\over6}\cr &=\textstyle{33\over54}.}

(All of this assumes, of course, equally likley outcomes with regards to which chip is chosen from bowl $A$, and with regards to which chip you chose from bowl $B$ afterwards.)

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