Hi there! I have a probability quiz this week and I grabbed some problems from my worksheet. This one in particular has me stumped:At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are lorries. If every vehicle is equally likely to leave, find the probability of:a) a van leaving first.b) a lorry leaving first.c) a car leaving second if either a lorry or van had left first.Can you help??

[tex]\begin{gathered} \text{Number of vehicles = 100} \\ \text{Number of cars =60} \\ Number\text{ of vans =30} \\ \text{Number of lorries =10} \end{gathered}[/tex]

Since each of the vehicles is equally likely to leave;

Part A

[tex]Pr(van)=\frac{\text{number of vans}}{Total\text{ number of vehicles}}=\frac{30}{100}=0.3[/tex]

Answer: 0.3

Part B

[tex]Pr(\text{lorry)}=\frac{number\text{ of lorries}}{\text{Total number of vehicles}}=\frac{10}{100}=0.1[/tex]

Answer: 0.1

Part C

First we find the probability of a lorry or van leaving

[tex]Pr(\text{Lorry or van) = }pr(lorry)+pr(Van)=0.1+0.3=0.4\text{ or }\frac{4}{10}[/tex]

Next, we find the probability of a car; but remember that one of either a lorry or van has left the car park already, so the total number of vehicles will reduce by 1

[tex]Pr(car)=\frac{60}{99}=\frac{20}{33}[/tex]

Therefore, the probability of a car leaving second if either a lorry or van had left first is

[tex]Pr((\text{lorry or van) and car)) }=\frac{4}{10}\times\frac{20}{33}=\frac{8}{33}[/tex]

Answer:

[tex]\frac{8}{33}[/tex]