For a loaded die, the probabilities of outcomes are given as under
$P(1) = P(2) = 0.2,$ $P(3) = P(5) = P(6) = 0.1$ and $P(4) = 0.3$.
The die is thrown two times. Let A and B be the events, 'same number each time' and 'a total score is 10 or more', respectively. Determine whether or not A and B are independent.

For a loaded die, it is given that
$P(1) = P(2) = 0.2$,

$P(3) = P(5) = P(6) = 0.1$ and $P(4) = 0.3$

Also, die is thrown two times.

Here, A = Same number each time and B = Total score is 10 or more

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

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

$= [P(1) \cdot P(1) + P(2) \cdot P(2) + P(3) \cdot P(3) + P(4) \cdot P(4) + P(5) \cdot P(5) + P(6) \cdot P(6)]$

$= [0.2 \times 0.2 + 0.2 \times 0.2 + 0.1 \times 0.1 + 0.3 \times 0.3 + 0.1 \times 0.1 + 0.1 \times 0.1]$

$= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20$
and $B = \{ (4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\}$

$\therefore$ $P(B) = P(4,6) + P(6,4) + P(5,5) + P(5,6) + P(6,5) + P(6,6)$

$= P(4) \cdot P(6) + P(6) \cdot P(4) + P(5) \cdot P(5) + P(5) \cdot P(6) + P(6) \cdot P(5) + P(6) \cdot P(6)$

$= 0.3 \times 0.1 + 0.1 \times 0.3 + 0.1 \times 0.1 + 0.1 \times 0.1 + 0.1 \times 0.1 + 0.1 \times 0.1$

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

Also, $A \cap B = \{ (5,5),(6,6)\}$

$\therefore$ $P(A \cap B) = P(5,5) + P(6,6) = P(5) \cdot P(5) + P(6) \cdot P(6)$

$= 0.1 \times 0.1 + 0.1 \times 0.1 = 0.01 + 0.01 = 0.02$

We know that, for two events A and B, if $P(A \cap B) = P(A) \cdot P(B)$,

then both are independent events.

Here, $P(A \cap B) = 0.02$ and $P(A) \cdot P(B) = 0.20 \times 0.10 = 0.02$

Thus, $P(A \cap B) = P(A) \cdot P(B) = 0.02$

Hence, A and B are independent events.