**Question **: A six-faced 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, the probability that the sum of two numbers thrown is even is

(A) 1/12

(B) 1/6

(C) 1/3

(D) 5/9

**Answer **: option (D)

**Solution **:

Let the probability for getting an odd number be p. Therefore, the probability for getting an even

number is 2p.

therefore, p+2p=1 ⇒ 3p=1 ⇒ p=1/3

Sum of two numbers is even means either both are odd or both are even. Therefore, the required

probability is,

\dfrac{1}{3}\times \dfrac{1}{3}+\dfrac{2}{3}\times \dfrac{2}{3}=\dfrac{1}{9}+\dfrac{4}{9}=\dfrac{5}{9}

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