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 = 0.10
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.
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}and n(6) and n(S) = 6 x 6 = 36
So, P(A) = n(A)/n(S) = 6/36 = 1/6
And, B = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}; n(B) = 6 and n(S) = 36
So, P(B) = n(B)/n(S) = 6/36 = 1/6
Now, A ⋂ B = {(5, 5), (6, 6)}
So, P(A ⋂ B) = 2/36 = 1/18
Hence, if A and B are not independent, then
P(A ⋂ B) = P(A).P(B)
1/18 ≠ 1/6 x 1/6 ⇒ 1/18 ≠ 1/36
Therefore, A and B are not independent events.
