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At a movie theater, the manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365-day year, what position in line gives you the best chance of being the first duplicate birthday?

please explain how you got this answer. Thank you!​

Sagot :

MrRoyal

The best chance is an illustration of probability

The 21st position in line gives you the best chance of being the first duplicate birthday

How to determine the best position?

Assume you are in the nth position, the probability of getting a free ticket is:

P(n) = P'(first n - 1 people share a birthday) * P(birthday with first n -1 people)

Mathematically, the above can be represented as:

P(n) = [365/365 * 364/365 * 363/365 * ... * (365 - (n -2))/365] * [(n -1)/365]

Such that: n [tex]\le[/tex] 365

When simplified, the equation becomes

p(n)/p(n+1) = 365/(366 - n) * (n - 1)/n

Where

p(n)/p(n+1) > 1

So, we have:

365/(366 - n) * (n - 1)/n > 1

Evaluate the product

(365n -365)/(366n - n²) > 1

Cross multiply

365n -365 > 366n - n²

Collect like terms

n² + 365n - 366n - 365 > 0

Evaluate the difference

n²  - n - 365 > 0

Solve for n using a graphing calculator;

n > - 18.6 or n >19.6

n cannot be negative.

So, we have:

n > 19.6

Approximate to nearest integer

n > 20

The least integer value of n is:

n = 21

Hence, the 21st position in line gives you the best chance of being the first duplicate birthday

Read more about probability at:

https://brainly.com/question/25870256

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