Given P(A) = 2/5 and P(A U B) = 7/10, find P(B) if A and B are mutually exclusive, and if they are independent.

The formula simplifies to $P(A \cup B) = P(A) + P(B)$.

The probability $P(B)$ when $A$ and $B$ are mutually exclusive is $\boxed{\frac{3}{10}}$.

Substitute this into the general formula: $P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$.

The probability $P(B)$ when $A$ and $B$ are independent is $\boxed{\frac{1}{2}}$.

8 (ii)
Let $X, Y, Z$ be the events that a truck is hired from Company X, Y, or Z, respectively.
Let $L$ be the event that a truck arrives late.

Given probabilities:
$P(X) = 40\% = 0.40$
$P(Y) = 50\% = 0.50$
$P(Z) = 10\% = 0.10$

Given conditional probabilities of arriving late:
$P(L|X) = 9\% = 0.09$
$P(L|Y) = 6\% = 0.06$
$P(L|Z) = 20\% = 0.20$

The probability that the truck will arrive late is $\boxed{0.086}$.

The probability that the truck will not arrive late is $\boxed{0.914}$.

Given that a truck hired arrives late, find the probability that it comes from Y.

Use Bayes' theorem:

$P(L)$ was calculated in part (c).

The probability that the truck comes from Y, given that it arrived late, is $\boxed{0.349}$.