Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
From which bowl are you drawing the chip? –  David Mitra Feb 2 '13 at 22:41
    
I am drawing from Bowl A –  Code Enthusiast 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. –  Code Enthusiast 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
show 3 more comments

1 Answer

up vote 1 down vote accepted

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.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.