DISCRETE MATH Consider the situation that you come to the movie theater booth an
ID: 3141767 • Letter: D
Question
DISCRETE MATH
Consider the situation that you come to the movie theater booth and are in the line for buying the movie ticket. There are 3 people in front of you and everyone in the line is supposed to buy at least one ticket. Assume the max number of tickets that one can buy is limited to 4 and the probability of the number of tickets that each person may buy is uniformly distributed.
(a) Suppose now there are only 5 tickets left. What is the probability that you can buy two tickets?
(b) Suppose now there are only 8 tickets left. What is the probability that you can buy two tickets?
Explanation / Answer
number of tickets one can buy = 1,2,3,4,
probability = 0.25 for each 1,2,3,4}
a) 5 tickets left, probability that you can buy two tickets = probability that 3 tickets is purchased by 3 person ahead =
P(X1 = 1 ,X2 = 2 ,X3 = 1) = 0.25^3 = 0.015625
b) there are only 8 tickets left. What is the probability that you can buy two ticket = P(X1 + X2 + X3 <= 6)
number of ways (X1,X2,X3) can be arranged =
X1 + X2 + x3 = n where X1> = 1
= (n-1)C2
X1+ X2 +X3 = 6 is 5C2 = 10 , {note this formula is applied as 4 is greater than maximum (4,1,1) }
X1+ X2 +X3 = 5 is 4C2 = 6
X1+ X2 +X3 = 4 3C2 = 3
X1+ X2 +X3 = 3 2C2 = 1
total ways = 10+6+3+1 = 20
hence required probabilty = 20*0.015625 = 0.3125
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