Given that for a loaded die-

P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3

And given that die is thrown two times and A is the event of same number each time

B is the event of a total score is 10 or more.

So,

A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

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

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

P(A)= [0.2×0.2+ 0.2×0.2+ 0.1×0.1+0.3×0.3+ 0.1×0.1+ 0.1

×0.1]

P(A)= [0.04+ 0.04+ 0.01+ 0.09+ 0.01+0.01]

P(A)= [0.20]

&

B= EVENT OF TOTAL SCORE IS 10 OR MORE

B= {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

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

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

P(B)= [0.3×0.1+ 0.1×0.1+ 0.1×0.1+ 0.1×0.3+ 0.1×0.1+ 0.1×0.1]

P(B)= [0.03+ 0.01+ 0.01+ 0.03+ 0.01+ 0.01]

P(B)= [0.10]

ALSO, probability of an intersection B (i.e. both the events occur simultaneously)

A Ո B = {(5,5), (6,6)}

HENCE,

P (A Ո B) = P (5,5) + P (6,6)

P (A Ո B) = P (5) × P (5) + P (6) ×P (6)

P (A Ո B) = 0.1×0.1+ 0.1×0.1

P (A Ո B) = 0.01+0.01

P (A Ո B) = 0.02

We know that if two events are independent than

P (A Ո B) = P(A) P(B)

HERE,

P(A). P(B)= 0.20× 0.10 = 0.02

SO, P (A Ո B) = P(A) P(B)

Hence, A and B are independent events.