Using the tree diagram from the previous answer, calculate the probability that the biscuit is: i) Made by machine J and is not broken.

This is the probability of a biscuit being made by Machine J AND not broken, denoted as $P(J \text{ and } B')$.

This is $P(B) = P(K \text{ and } B) + P(J \text{ and } B) + P(L \text{ and } B)$.
From the tree diagram:
$P(K \text{ and } B) = 0.45 \times 0.03 = 0.0135$
$P(J \text{ and } B) = 0.30 \times 0.02 = 0.0060$
$P(L \text{ and } B) = 0.25 \times 0.05 = 0.0125$

This means the biscuit is broken AND it was made by machine J OR machine L.

From the tree diagram:
$P(J \text{ and } B) = 0.0060$
$P(L \text{ and } B) = 0.0125$

We know $\vec{BO} = -\vec{OB}$ and $\vec{OB} = \mathbf{b}$.
So, $\vec{BO} = -\mathbf{b}$.
We are given $\vec{OA} = \mathbf{a}$.