Correct option b. (20/33)
Explanation:
Let S be the sample space and A be the event of a van leaving first. n(S) = 100 , n(A) = 30
Probability of a van leaving first: P(A) = 30/100 = 3/10
Let B be the event of a lorry leaving first. n(B) = 100 – 60 – 30 = 10
Probability of a lorry leaving first: P(B) = 10/100 = 1/10
If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars. Let T be the sample space and C be the event of a car leaving second.
n(T) = 99, n(C) = 60
Probability of a car leaving after a lorry or van has left:
P(B) = 60/99 = 20/33.