Question 1192024: Cinema seats are numbered with those numbers: 1, 2, ..., 20. 6 friends one after another buys tickets and seats are allocated to them at random order.
a) What are the chances that everyone will get seats with even numbers?
b) 2 friends that come earlier got seats with even numbers. What are the probability that the rest will also get places with even numbers?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
(a) At the start, there are 10 even-numbered seats out of a total of 20.
P(1st seat # is even) = 10/20
P(2nd seat # is even) = 9/19
...
P(5th seat # is even) = 6/16
P(6th seat # is even) = 5/15
You want all those things to happen, so multiply all those probabilities. I leave it to you to do the calculations and express the result in the required format.
(b) In this case, 2 of the 10 even-numbered seats are already occupied. The solution process is the same as in part (a); but now the probabilities are 8/18, 7/17, ..., 4/14, and 3/13.
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