2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you figure out the first term? What cases are there to distinguish given that we have heads? What are their probabilities? $\endgroup$ – Lord_Farin Apr 10 '13 at 11:10
  • 1
    $\begingroup$ @Lord_Farin Ok, it should 1/2 * (3/5 + (2/5) * 1/2). $\endgroup$ – Dude Apr 10 '13 at 11:16
  • 1
    $\begingroup$ Yes, that's correct. A similar case distinction (three cases) will get you the second term. $\endgroup$ – Lord_Farin Apr 10 '13 at 11:21
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $$1/2*3/5 + 1/2*2/5*1/2 +(1/2*2/5*1/4*1/3) + (1/2*3/5*2/4*2/3) + (1/2*3/5*2/4*1) = 2/3$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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