Question

It is known that a bus will arrive at random at a certain location sometime between 3:00 P.M. and 3:30 P.M. A man decides that he will go at random to this location between these two times and will wait at most 5 minutes for the bus. If he misses it, he will take the subway. What is the probability that he will take the subway?

Comments

@arjun is the ans 0.8472 ??

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I got 0.85 :P

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how to got .8472

Answer 1

the white area is our desire outcome and the total area is total outcome .

(0.5*25*25+30*30*0.5)/30*30=0.8472

Comments

How did you draw this diagram..please explain. Is it any theorom?

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Nopes. Just the representation of the fact in question. The diagonal line represents the time at which the man arrives. He catches the bus if it arrives within 5 minutes. So, the line for the bus is shifted by 5 in the x axis and has the same slope.

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Can also go like this if figure is not intuitive,

The bus and man can arrive in any time between 3:00 and 3:30.

Probability of man arriving before 3:25 = 5/6 as each 5 minute interval is equally possible for his arrival.

Probability of man arriving before 3:25 and catching the bus = 5/6 * 1/6 = 5/36 (as the probability for bus to come in any 5 minute interval is 1/6).

Probability of man to come after 3:25 = 1/6

Probability of man coming after 3:25 and catching the bus = 1/6 * 1/6 *x = x/36, where x is the probability that man comes first. Since both man and bus are equally likely to come first, x = 1/2, giving required probability = 1/72.

So, Probability that man catches the bus = 5/36 + 1/72 = 11/72.

Probability that he takes subway = 1 - 11/72 = 61/72 = 0.8472.

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@Arjun sir, what kind of distribution is this ..?? where can i read about it..??

Answer 2

The bus and man can arrive in any time between 3:00 and 3:30.

Probability of man arriving before 3:25 = 5/6 as each 5 minute interval is equally possible for his arrival.

Probability of man arriving before 3:25 and catching the bus = 5/6 * 1/6 = 5/36 (as the probability for bus to come in any non-overlapping 5 minute interval- 0-5, 1-6...  is 1/6 as probability for bus arrival must be 1/6 for every consecutive 5 minutes out of 30 minutes).

Probability of man to come after 3:25 = 1/6

Probability of man coming after 3:25 and catching the bus = 1/6 * 1/6 *x = x/36, where x is the probability that man comes first. Since both man and bus are equally likely to come first, x = 1/2, giving required probability = 1/72.

So, Probability that man catches the bus = 5/36 + 1/72 = 11/72.

Probability that he takes subway = 1 - 11/72 = 61/72 = 0.8472.

Comments

Why isn't the probability for the first 5 intervals where the man catches the bus something like

5*(1/6* 1/6* 1/2)? Because even in the first 5 intervals, the man has to arrive first.

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yes, but the thing is if man arrives at say 3:07, bus cane come between 3:07-3:12- a 5 minute interval. But in the last 5 minute interval, bus must come strictly after the arrival of man as there is no possibility of bus coming after 3:30.

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Thank you for clearing my doubt sir :)