Question 1200742: The free seats in a movie theater are numbered 1, 2, ..., n. m friends buy tickets one after the other, and seats are assigned to them randomly. What is the probability that they will all get seats with even numbers? Previously, k friends received seats with even numbers. What is the probability that the remaining friends will also get seats with even numbers? ([n, m, k] = [33, 6, 4])
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! Certainly, let's calculate the probability of all friends getting seats with even numbers given the provided information.
**1. Determine the number of even-numbered seats:**
* Total number of seats (n) = 33
* Number of even-numbered seats = n // 2 = 33 // 2 = 16
**2. Calculate the probability for the remaining friends:**
* Number of friends who already have even seats (k) = 4
* Number of remaining friends = m - k = 6 - 4 = 2
* Remaining even-numbered seats = 16 - 4 = 12
* Remaining total seats = 33 - 4 = 29
* Probability of the first remaining friend getting an even seat: 12/29
* Probability of the second remaining friend getting an even seat: 11/28
* Probability of both remaining friends getting even seats: (12/29) * (11/28) ≈ 0.162562
**Therefore, the probability that all friends will get seats with even numbers, given that 4 friends already have even seats, is approximately 0.162562.**
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