Your company operates a van service from the airport to downtown hotels. Each va

Question

Your company operates a van service from the airport to downtown hotels. Each van carries seven passengers. Many passengers who reserve seats don't show up-in fact, the probability is 0.3 that a randomly chosen passenger will fail to appear. Passengers behaviors are independent. If you allow nine reservations for each van, what is the probability that more than seven passengers will appear? Do a simulation to estimate this probability. (Using line 102 of this table of random digits, simulate 10 vanloads. Let 0-2 be a passenger failing to appear and 3-9 be a passenger appearing.)

Explanation / Answer

The first 9 random digits in row 102 are : 7,3,6,7,6,4,7,1,5

Note that out of these 9 digits only one is less than 2. Which means that 8 passengers are going to show up on this vanload.

Similarly, we can run the simulation of 9 other vanloads using the same row as shown -

As we can see that 9 out of 10 simulations have not more than 7 passenges showing up, so probability that more than 7 passengers will show up is = 0.1.

Passenger Random Digit Appear (Y / N) More than 7 Passengers (Y/N) Vanload 1 1 7 1 Y 2 3 1 3 6 1 4 7 1 5 6 1 6 4 1 7 7 1 8 1 0 9 5 1 Vanload 2 1 0 0 N 2 9 1 3 9 1 4 4 1 5 0 0 6 0 0 7 0 0 8 1 0 9 9 1 Vanload 3 1 2 0 N 2 7 1 3 2 0 4 7 1 5 7 1 6 5 1 7 4 1 8 4 1 9 2 0 Vanload 4 1 6 1 N 2 4 1 3 8 1 4 8 1 5 2 0 6 4 1 7 2 0 8 5 1 9 3 1 Vanload 5 1 6 1 N 2 2 0 3 9 1 4 0 0 5 7 1 6 3 1 7 6 1 8 7 1 9 6 1 Vanload 6 1 4 1 N 2 7 1 3 1 0 4 5 1 5 0 0 6 9 1 7 9 1 8 4 1 9 0 0 Vanload 7 1 0 0 N 2 0 0 3 1 0 4 9 1 5 2 0 6 7 1 7 2 0 8 7 1 9 7 1 Vanload 8 1 5 1 N 2 4 1 3 4 1 4 2 0 5 6 1 6 4 1 7 8 1 8 8 1 9 2 0 Vanload 9 1 4 1 N 2 2 0 3 5 1 4 3 1 5 6 1 6 2 0 7 9 1 8 0 0 9 7 1 Vanload 10 1 3 1 N 2 6 1 3 7 1 4 6 1 5 4 1 6 7 1 7 1 0 8 5 1 9 0 0

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