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Mathematics

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Assume there are 365 days in a year.

babunello [11817]

Response:

1) The likelihood that ten students in a classroom have distinct birthdays is 0.883.

2) The likelihood that at least two out of ten students share a birthday is 0.002.

Detailed explanation:

Given: Assuming 365 days are in a year.

To find: 1) What is the probability that ten students in a classroom have unique birthdays?

2) What is the probability that at least two among ten students have a birthday in common?

Solution:

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

Total outcomes = 365

1) The chance that ten students in a class hold different birthdays is

The first student can have their birthday on any of the 365 days, the second can only on 364/365 and so forth...

$\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times \frac{361}{365}\times\frac{360}{365} \times \frac{359}{365} \times \frac{358}{365} \times \frac{357}{365} \times\frac{356}{365}=0.883$

The chance that ten students in a class have distinct birthdays is 0.883.

2) The likelihood that at least two out of ten students share a birthday

P(2 on the same day) = 1 - P(2 not on the same day)

$\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]$

$\text{P(2 born on same day) }=1-[\frac{364}{365}]$

$\text{P(2 born on same day) }=0.002$

The probability that at least two students in a class have the same birthday is 0.002.

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