1. May 17, 2012

### Whitebunny

1. The problem statement, all variables and given/known data
Ten balls are in an urn; five red and five blue. When a red ball is removed, it is always
replaced. When a blue ball is removed a coin is flipped. If heads appears the ball is not
replaced. If tails appears, then two blue ones are placed into the urn. What is the
probability that the first three balls drawn are the same color?

2. Relevant equations
I honestly have no idea. Very sorry.

3. The attempt at a solution
Draw 1
Red 50% chance, Blue 50% (heads 50%, tails 50%)

Draw 2 (this is where I get confused)
if first draw red then, Red 50% chance, Blue 50% (heads 50%, tails 50%)
if first draw blue and tails then, Red 5/11, blue 6/11 (heads 50%, tails 50%)
If first draw blue and heads then, Red 5/9, blue 4/9 (heads 50%, tails 50%)

Draw 3
if first draw red, if second draw red then, Red 50% chance, Blue 50%
if first draw red then, and second draw is blue and tails then, Red 5/11, Blue 6/11
If first draw red, and second draw is blue and heads then, Red 5/9, Blue 4/9
If first draw is blue and tails, and second draw is red, then Red 5/11, Blue 6/11
If first draw is blue and tails, and second draw is blue and tails, then Red 5/12, Blue 7/12
If first draw is blue and tails, and second draw is blue and heads, the Red 50% chance, Blue 50%
If first draw is blue and heads, and second draw is red, then Red 5/9, Blue 4/9
If first draw is blue and heads, and second draw is blue and tails then, the Red 50% chance, Blue 50%
If the first draw is blue and heads, and the second draw is blue and heads, then Red 5/8 chance, Blue 3/8
Then I multiply everything together (?)
.5*.5*.5*.5*5/11*6/11*5/9*4/9*.5*.5*5/11*6/11*5/9*4/9*5/12*7/12*.5*.5*5/9*4/9*.5*.5*5/8*3/8
=itty bitty number that can’t possible be the answer.

Last edited: May 17, 2012
2. May 17, 2012

### HallsofIvy

Set up a tree:

R B
/ \ / \
R B R B
/ \ / \ / \ / \
R B R B R B R B

Calculate the probability, step by step, of each of those 8 outcomes.

3. May 17, 2012

### Whitebunny

Are you sure this works here? Because of the coin flipping and removal, shouldn't there be more possible outcomes? Could you please elaborate a bit more? I really don't understand...

Last edited: May 17, 2012
4. May 17, 2012

### Ray Vickson

Try to do it systematically. Let's look at the contents of the urn *after* the first event.
$$\begin{array}{ccc} \text{first event}&\text{probability} & \text{new contents}\\ \hline \text{red} & 1/2 & \text{5 red, 5 blue}\\ \text{blue,heads} & 1/4 &\text{5 red, 4 blue}\\ \text{blue,tails}& 1/4 & \text{5 red, 6 blue} \end{array}$$
So, before drawing the second ball the urn has either 5 red and 5 blue, or 5 red and 4 blue, or 5 red and 6 blue, and you have probabilities of each. Now you can look at what happens after drawing the second ball, etc., etc. Essentially, there will always be 5 red balls, so you can just look at the number of blue balls at each stage. If you have studied Markov chains, this would be an obvious example; if not, you can just do it manually. It really is not too bad: you are asked to find the probability that all three balls drawn are the came color. So the drawings are either RRR or BBB. Getting P{RRR} should be easy. Getting P{BBB} needs a bit more work, along the lines of what I showed above. In getting P{BBB} the color is always blue, so the only factors you need worry about are the first two coin-toss results; you should ask yourself whether HT and TH would each give the same answer.

RGV

Last edited: May 17, 2012