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The 3.4.a can be handled normally because it is all about the probability of the first red roll. However, the 3.4.b is able to be complicated easily because what we have to deal with is the conditional probability in the second roll. | The 3.4.a can be handled normally because it is all about the probability of the first red roll. However, the 3.4.b is able to be complicated easily because what we have to deal with is the conditional probability in the second roll. | ||
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Eventually the answer could be | Eventually the answer could be | ||
Latest revision as of 07:07, 15 September 2008
The 3.4.a can be handled normally because it is all about the probability of the first red roll. However, the 3.4.b is able to be complicated easily because what we have to deal with is the conditional probability in the second roll.
The second red roll is given first red roll. The first red roll would be 1/3.
What I think at this point is that I must calculate as each possibility of second roll times first red roll probability.
such as
(1/3)*{ (1/2)*(1/6)+(1/2)*(1/2)} + (1/2)*{(1/2)*(1/6)+(1/2)*(1/3)} + (1/6)*{(1/2)*(1/3)+(1/2)*(1/2)}
Eventually the answer could be
P(second red roll | First red roll) <- The answer
I wish that this is correct.
If you think it is wrong or has any troubles, Plz give me comments.
