For a loaded die, the probabilities of outcomes are given as under:

Solution:

Given that a loaded die is thrown such that

P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3 and die is thrown two times. Also given that:

A = same number each time and

B = Total score is 10 or more.

So, P(A) = [P(1, 1) + P(2, 2) + P(3, 3) + P(4, 4) + P(5, 5) + P(6, 6)]

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

= 0.2 x 0.2 + 0.2 x 0.2 + 0.1 x 0.1 + 0.3 x 0.3 + 0.1 x 0.1 + 0.1 x 0.1

= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20

Now, B = [(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)]

P(B) = [P(4).P(6) + P(6).P(4) + P(5).P(5) + P(5).P(6) + P(6).P(5) + P(6).P(6)]

= 0.3 x 0.1 + 0.1 x 0.3 + 0.1 x 0.1 + 0.1 x 0.1 + 0.1 x 0.1 + 0.1 x 0.1

= 0.03 + 0.03 + 0.01 + 0.01 + 0.01 + 0.01 = 0.10

A and B both events will be independent if

P(A ⋂ B) = P(A).P(B) …. (i)

And, here (A ⋂ B) = {(5, 5), (6, 6)}

So, P(A ⋂ B) = P(5, 5) + P(6, 6) = P(5).P(5) + P(6).P(6)

= 0.1 x 0.1 + 0.1 x 0.1 = 0.02

From equation (i) we get,

0.02 = 0.20 x 0.10

0.02 = 0.02

Therefore, A and B are independent events.