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A six faces dice is so biased that it is twice as likely to show an even number as an odd number when thrown.It is thrown twice. What is the probability that the sum of two numbers thrown is even.

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    $\begingroup$ P = P(odd odd) + P(even even) $\endgroup$ – Cato Jan 6 '17 at 11:43
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The sum is even if both throws are even or both odd. Since now there is a $2/3$ chance of an even number each throw, the combined probability is $(2/3)^2 + (1/3)^2 = \mathbf{5/9},$ which is more than the $(1/2)^2 + (1/2)^2 = 1/2,$ that would be expected for unbiased dice.

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so if you make P(EVEN) = A

P(ODD) = B

A = 2B

A + B = 1 (by total probability is always 1)

so turns out A = 2/3

P = P(odd odd) + P(even even) = (1/3) x (1/3) + (2/3) x (2/3) = 5 / 9

answer 5/9

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Exclude the cases where we have one even and one odd number. So considering the two different ordres the desired probability is $$1-\left[\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)\left(\frac{2}{3}\right) \right]$$

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P(showing even no)=2/(1+2)=2/3 P(showing odd no)=1/(1+2)=1/3 For Sum to be odd=(E+E+O)or(E+O+E)or(O+E+E)or(O+O+O) =(2/3*2/3*1/3)*3+(1/3*1/3*1/3)=13/27 this will be the answer.

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    $\begingroup$ Please, use MathJax (i.e. LaTeX commands) for mathematical notations. $\endgroup$ – Taroccoesbrocco Aug 21 '18 at 5:44

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