Question

Consider the following case: Buses arrive at a particular bus stop after every 15 minutes, starting from 6AM. If a passenger arrives at the stop at a random time which is uniformly distributed between 9 am to 9.30 am, then find the probability that he waits for (a) less than 5 minutes for a bus (b) atleast 10 minutes for a bus

Answer 1

Answer on Question #60972 – Math – Statistics and Probability

Question

Consider the following case: Buses arrive at a particular bus stop after every 15 minutes, starting from 6AM. If a passenger arrives at the stop at a random time which is uniformly distributed between 9 am to 9.30 am, then find the probability that he waits for

(a) less than 5 minutes for a bus

(b) at least 10 minutes for a bus

Solution

Let's draw the time axis, where the origin (0) corresponds to 9 am:

Hatches indicate time when bus arrives. Thick hatches specify the period [0; 30] when a passenger arrives at the stop.

a) Blue regions ((10; 15], (25; 30]) indicate areas that are less than 5 minutes before bus arrival. Their spread is 15-10=5 mins.

By geometric definition of probability, in that case $P = \frac{S_{blue}}{S_{total}} = \frac{5 + 5}{30} = 1/3$.

b) Orange regions ([0; 5), [15; 20]) indicate areas that are more than 10 minutes before bus arrival. Their spread is 20-15=5 mins, therefore, the probability in that case is

$$P = \frac{S_{orange}}{S_{total}} = \frac{5 + 5}{30} = 1/3.$$

Answer: a) 1/3; b) 1/3.

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