In the city 2

In the city Náhoda (Coincidence), they abolished the timetables and the city buses run completely randomly. Imagine that you are standing at a bus stop, at which 5 buses stop with numbers 12, 14, 15, 21 and 27, and you can get home by two of them.
a/What is the probability that you can get home right by the first bus that stops?
b/What is the probability that you cannot get home right by the first bus that stops?
c/What is the probability that as the first there comes a bus whose number is divisible by three?

Final Answer:

p₁ =  0.4
p₂ =  0.6
p₃ =  0.8

Step-by-step explanation:

$p_1=\frac{2}{5}=0.4$

$p_2=1-p_1=1-\frac{2}{5}=\frac{1\cdot 5}{5}-\frac{2}{5}=\frac{5}{5}-\frac{2}{5}=\frac{5-2}{5}=\frac{3}{5}=0.6$

$p_3=\frac{4}{5}=0.8$

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