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An urn contains 3 Red and 4 White marbles. A fair coin is flipped. If the flip is Heads then 1 Red and 2 White marbles are added to the urn. On the other hand, if the flip is Tails, then 1 Red and 2 White marbles are removed from the urn. Two random marbles are now drawn from the urn without replacement. (i) What is the probability that both of the drawn marbles are White? (ii) What is the probability that the flip was Heads, given that the two drawn marbles have different colors?

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1 Answer 1

Consider the result of the coin flip. If it is heads, then:

$$ \text{No. of red marbles}_H = 3+1 = 4\\ \text{No. of white marbles}_H = 4+2 = 6\\ $$

If it is tails, then:

$$ \text{No. of red marbles}_T = 3-1 = 2\\ \text{No. of white marbles}_T = 4-2 = 2\\ $$

In the heads situation, $$ P(\text{both marbles are white}) = \frac{6\choose2}{10 \choose 2} $$

Similarly, in the tails situation, $$ P(\text{both marbles are white}) = \frac{2\choose2}{4 \choose 2} $$

You will have to multiply both values by $0.5$ to take into account the result of the coin toss.

(b) For the second part, the answer is $$ P(\text{1 red & 1 white, and coin flip was heads}) = 0.5 \times \frac{{4\choose 1}{6\choose 1}}{10 \choose 2} $$

$$ P(\text{1 red & 1 white}) = 0.5 \times \frac{{4\choose 1}{6\choose 1}}{10 \choose 2} + 0.5 \times \frac{{2\choose 1}{2\choose 1}}{4 \choose 2} $$

You can use the conditional probability formula to get the answer.

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in the tails situation No. of white marblesT= 4−2=2 so this should give me (2C2) and shouldn't the tails situation be (2C2)/(4C2)? –  reddoor Feb 21 '13 at 3:32
    
yep, i'll edit to reflect the changes. –  bryansis2010 Feb 21 '13 at 3:40
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