Question

Assume there are 365 days in a year.

Answer 1

Answer:

1) The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.

Step-by-step explanation:

Given : Assume there are 365 days in a year.

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

2) What is the probability that among ten students in a class, at least two of them share a birthday?

Solution :

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

Total outcome = 365

1) Probability that ten students in a class have different birthdays is

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

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

The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday

P(2 born on same day) = 1- P( 2 not born on 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 among ten students in a class, at least two of them share a birthday is 0.002.