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## Question

A boy is late for school often. When approached by his teacher, he explained that it is not his fault. Then he provided some details. His father takes him from home to the bus stop every morning. The bus is supposed to leave at 8:00 am but the departure time is only approximate. The bus arrives at the stop anytime between 7:58 and 8:02 and immediately departs. The boy and his father try to arrive at the bus stop at 8:00 however due to variable traffic conditions they arrive anytime between 7:55 and 8:01. This is why the boy misses the bus so often.

Can you determine how often the boy is late for school?

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• 0

Is the arrival time continuous or discrete (whole minutes)?

Can we assume the distribution is uniform?

Continuous time is the interesting case so let's look at that with uniform distribution:

Let t=0 be 7:58, so the bus arrives anywhere in time [0,4], the kid arrives anywhere in [-3,3].

50% chance the kid arrives in segment [-3,0] and in this case his chance of catching the bus is 1.

50% chance the kid arrives in segment [0,+3] at time t1, then he would catch the bus iff it arrives in the interval [t1,4], so his chance is (4-t1)/4 = 1-t1/4, so on average that's the integral from 0 to 3 of 1-x/4 divided by 3 for normalization, which is (3-9/8)/3 = 1-3/8 = 5/8

So the final probability is (1+5/8)/2 = 13/16

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(1) If the boy arrives between 7:55 and 7:58, then his chance to catch the bus equals 1.
(2) If the boy arrives between 7:58 and 8:01 and the buss arrives between 7:58 and 8:01, then his chance to catch the bus equals 1/2.
(3) If the boy arrives between 7:58 and 8:01 and the buss arrives between 8:01 and 8:02, then his chance to catch the bus equals 1.

Case (1) has probability 1/2, case (2) - 1/2 * 3/4, and case (3) - 1/2 * 1/4,
therefore the boy's chance to catch the bus equals
1/2 * 1 + 1/2 * 3/4 * 1/2 + 1/2 * 1/4 * 1 = 8/16 + 3/16 + 2/16 = 13/16

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witzar has it IMO

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I think witzar is close

if the boy has an even chance of arriving at any minute between 7:55 and 8:02, then every minute has a 1/8 probability

if the bus arrival is also evenly distributed between 7:58 and 8:02, then every minute has a 1/5 probability

If the boy arrives at

7:55 the probability of catching the bus is 1.0

7:56 the probability of catching the bus is 1.0

7:57 the probability of catching the bus is 1.0

7:58 the probability of catching the bus is 1.0

7:59 the probability of catching the bus is 4/5

8:00 the probability of catching the bus is 3/5

8:01 the probability of catching the bus is 2/5

8:02 the probability of catching the bus is 1/5

so the probability that he catches the bus is

1/8* [1+1+1+1+ 0.8 + 0.6 + 0.4 + 0.2] = 1/8 * 6 = 3/4 = 75%

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I think witzar is close

if the boy has an even chance of arriving at any minute between 7:55 and 8:02, then every minute has a 1/8 probability

if the bus arrival is also evenly distributed between 7:58 and 8:02, then every minute has a 1/5 probability

If the boy arrives at

7:55 the probability of catching the bus is 1.0

7:56 the probability of catching the bus is 1.0

7:57 the probability of catching the bus is 1.0

7:58 the probability of catching the bus is 1.0

7:59 the probability of catching the bus is 4/5

8:00 the probability of catching the bus is 3/5

8:01 the probability of catching the bus is 2/5

8:02 the probability of catching the bus is 1/5

so the probability that he catches the bus is

1/8* [1+1+1+1+ 0.8 + 0.6 + 0.4 + 0.2] = 1/8 * 6 = 3/4 = 75%

The OP does not constrain the boy's arrival to be on the minute.

Between 7:58 and 7:59, for example, p decreases linearly from 1 to 0.8; it does not drop abruptly to 0.8 at 7:58.

Similarly for the next two minute-intervals.

The boy does not arrive later than 8:01.

Make a graph of p vs boy's arrival time and find its average value.

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I think witzar is close

if the boy has an even chance of arriving at any minute between 7:55 and 8:02, then every minute has a 1/8 probability

if the bus arrival is also evenly distributed between 7:58 and 8:02, then every minute has a 1/5 probability

If the boy arrives at

7:55 the probability of catching the bus is 1.0

7:56 the probability of catching the bus is 1.0

7:57 the probability of catching the bus is 1.0

7:58 the probability of catching the bus is 1.0

7:59 the probability of catching the bus is 4/5

8:00 the probability of catching the bus is 3/5

8:01 the probability of catching the bus is 2/5

8:02 the probability of catching the bus is 1/5

so the probability that he catches the bus is

1/8* [1+1+1+1+ 0.8 + 0.6 + 0.4 + 0.2] = 1/8 * 6 = 3/4 = 75%

The OP does not constrain the boy's arrival to be on the minute.

Between 7:58 and 7:59, for example, p decreases linearly from 1 to 0.8; it does not drop abruptly to 0.8 at 7:58.

Similarly for the next two minute-intervals.

The boy does not arrive later than 8:01.

Make a graph of p vs boy's arrival time and find its average value.

good point!

In fact the boy does not necessarily arrive on the minute and the bus does not necessarily arrive on the minute either.

I think your approach is better.

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