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Sagot :
Explanation
In the question, we are given that;
[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]
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