a) The probabilities at each fork have to add up to 1.
At the first light, if P(not green) = 0.2, then P(green) = 1 - 0.2 = 0.8.
If the first light is green, and at the second light P(green) = 0.6, then P(not green) = 0.4.
If the first light is not green, and at the second light P(green) = 0.3, then P(not green) = 0.7.
b) The probability that the first light is green is 0.8.
c) The first light is not green, so we take the bottom path at the first fork, where P(not green) = 0.2. Then the probability that the second light is green is 0.3.
d) The probabilities at the second set of lights are conditional probaiblities. For example, 0.6 refers to the probability that the second light is green *on the condition that* the first light is green.
By definition of conditional probability,
P(first green and second green) = P(second green | first green) P(first green)
and we have
P(second green | first green) = 0.6
P(first green) = 0.8
so that
P(first green and second green) = 0.6 • 0.8 = 0.48
e) The event that at least one set is green involves 3 possible cases:
• both lights are green
• first light is green, second light is not
• first light is not green, second light is
These events are mutually exclusive - they can't all happen at the same time - so the total probability is the sum of the individual probabilities.
• Case 1:
We know from part (d) that both lights are green with probability 0.48.
• Case 2:
From our tree diagram, we find
P(second not green | first green) = 0.4
P(first green) = 0.8
⇒ P(first green and second not green) = 0.4 • 0.8 = 0.32
• Case 3:
The diagram tells us that
P(second green | first not green) = 0.3
P(first not green) = 0.2
⇒ P(first not green and second green) = 0.3 • 0.2 = 0.06
Then the total probability that at least one light is green is
P(at least 1 green) = 0.48 + 0.32 + 0.06 = 0.86
