A biased die is twice as likely to show an even number as an odd number.  The die is rolled three times.  if occurance of an even number is considered a success, then write the probability distribution of number of successes.

Probability of success in one throw (p) = 2/3

So probability of getting r success in n throws = ⁿC_r*p^r*(1-p)^n-r = ⁿC_r*(2/3)^r*(1/3)^n-r 

Let p be the prob.  of getting an odd no. in a single throw of a die. Then the prob. of getting an even no. is 2p.

We have,

P(1) + P(2) + P(3) + P(4) +P(5) + P(6) = 1

or, p + 2p + p + 2p + p + 2p = 1

or, p = 1/9

So, prob. of getting an even no. = 2/9 and prob. of getting an odd no. = 1/9.

Since an even no. is considered as success,

Let X = no. of even no. on 2 throws.

When X = 0, P(X) = (1/9)(1/9)(1/9) = 1/729

When X = 1, P(X) = (1/9)(1/9)(2/9)(3) = 6/729 = 2/243

When X = 2, P(X) = (1/9)(2/9)(2/9)(3) = 4/243

When X = 3, P(X) = (2/9)(2/9)(2/9) = 8/729

whose ans is correct.

I think its Deepak who is correct.

yes deepak is totally correct....