# What is the probability in this case?

Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls, and $U_2$ contains only $1$ white ball.

A fair coin is tossed:

• If head appears then $1$ ball is drawn at random from $U_1$ and put into $U_2$.
• If tail appears then $2$ balls are drawn at random from $U_1$ and put into $U_2$.

Now $1$ ball is drawn at random from $U_2$.

Question is:

What is the probability of the drawn ball from $U_2$ being white?

Ok I so don't have much clue about this question. I came across this question in one of my tests. This is what I have

$$\Pr(\text{Ball is 'white'}) = {\Pr(\text{Ball is 'white'} \mid \text{heads}) \over \Pr(\text{heads})} + {\Pr(\text{Ball is 'white'} \mid \text{tails}) \over \Pr(\text{tails})}$$

But I can't figure out how to calculate $\Pr(\text{Ball is 'white'})$. I have the answer too, but it is not intuitive enough so I can't deduce what term refers to what probability.

Edit: Question has been updated incorrectly, Pr(Heads) so is Pr(tails) should be a nominator rather than being a denominator.

• Can you figure out the first term? What cases are there to distinguish given that we have heads? What are their probabilities? – Lord_Farin Apr 10 '13 at 11:10
• @Lord_Farin Ok, it should 1/2 * (3/5 + (2/5) * 1/2). – Dude Apr 10 '13 at 11:16
• Yes, that's correct. A similar case distinction (three cases) will get you the second term. – Lord_Farin Apr 10 '13 at 11:21

Sol:- case 1:- head, white from U1, white from U2= (1/2)(3/5)(2/2)= 3/10 Case2:- head, red from U1, white from U2= (1/2)(2/5)(1/2)1/10 Case 3:- tail, 2 white from U1, white from U2= (1/2)(3C2/5C2)(3/3)= 3/20 Case 4:-P( tail, white and red from U1, white fromU2)= (1/2)(3C1*2C1)/(5C2) (2/3)=1/5 case 5:- P(tail, 2 red fom U2, white from U1)= (1/2)(2C2/5C2)(1/3)= 1/60 Required Probability= sum of probabilities of above cases= (3/10)+ (1/10)+(3/20)+(1/5)+(1/60)= 46/60= 23/30

Think about how many balls will be in U2 under different scenarios

1. Get H and then choose a white ball; then have two W balls in U2: probability of that happening is $$1/2*3/5*1$$. OR Get H and choose R ball. Then have 1 W out of 2 balls in U2. Probability of choosing a W from U2 is now $$1/2*2/5*1/2$$

2. Get T. Now will add 0,1 or 2 W balls to U2.

2.1 Add 0 W to U2. Then have 1 W out of 3 in U2. Probability of drawing W from U2 is then $$1/2*2/5*1/4*1/3$$

2.2 Add 1 W to U2. Then have 2 W out of 3 in U2. Probability of drawing W from U2 is then $$1/2*3/5*2/4*2/3$$

2.3 Add 2 W to U2. Then have 3 W out of 3 in U2. Probability of drawing W from U2 is then $$1/2*3/5*2/4*1$$

So total probability is $$1/2*3/5 + 1/2*2/5*1/2 +(1/2*2/5*1/4*1/3) + (1/2*3/5*2/4*2/3) + (1/2*3/5*2/4*1) = 2/3$$