Question

A die is manufactured in such a way that an even number is twice likely to occur as an odd number. If the die is tossed twice, find the probability distribution of the random variable X representing the perfect square in the two tosses.

Answer 1

Let X represents the number of perfect squares. 

S = sample space = { 1, 2, 3. 4, 5, 6} 

∴ n(S) 6 

n(X) = Possible number of perfect squares = {1,4} = 2 

∴ Probability of getting perfect square

= n(X)/n(S) = 2/6

And probability of not getting perfect square

= 1 - 2/6 = 4/6

When die is tossed twice n(S) = 6 x 6 = 36

∴ P(X = 0) = 8 (No perfect square)

= 4/6 x 4/6 = 4/9

P(X = 1) = P(one perfect square)

= 2/6 x 4/6 + 4/6 x 2/6

= 2/9 + 2/9 = 4/9

P(X = 2) = P (both are perfect square)

∴ P(X = 2) = 2/6 x 2/6 = 1/9

∴ Probability distribution is following: