Joint, total and reverse conditional probabilities from a tree diagram, and the distribution and mean of a two-day cost

On any working day, the probability that a delivery van is held up in traffic is \(\frac25\). When the van is held up, the probability that its delivery is late is \(\frac34\). When it is not held up, the probability that its delivery is late is \(\frac14\). Let \(T\) be the event that the van is held up in traffic and \(L\) the event that the delivery is late. The tree diagram below shows this information.

1. Find
   1. \(\mathrm{P}(T\cap L)\)
   2. \(\mathrm{P}(L)\)[4]
2. Find the probability that
   1. the van is held up in traffic and the delivery is on time
   2. the van was held up in traffic, given that the delivery is late.[5]

On each day that the van is held up in traffic, it uses extra fuel costing \(6\) dollars. The van works on a Thursday and on the following Friday, and the two days are independent. Let \(X\) be the total cost, in dollars, of extra fuel over these two days.

3. Copy and complete the following probability distribution table for \(X\).[3]

   \(x\)	\(0\)	\(6\)	\(12\)
   \(\mathrm{P}(X=x)\)	\(\frac{9}{25}\)

4. Find the expected total cost of extra fuel over the two days.[2]