365 days in a year.

1) what is the probability that ten students in a class have different birthdays?
2) what is the probability that at least ten students in a class share their birthday?

1) What is the probability that ten students in a class have different birthdays?
We know in one year we have 365 days (omitting leap years)

The first students can have his birthday on any day so:

P(1st student)= 365/365 = 1 (assertion)
P(2nd students) = 364/365
P(3rd students) = 363/365
P(4th students) = 362/365
P(5th students) = 361/365
P(6th students) = 360/365
P(7th students) = 359/365
P(8th students) = 358/365
P(9th students) = 357/365
P(10th students) = 356/365
Then the probability that none of the birthdays happen at the same times:
365/365 x 364/365 x 363/3654 ...357/365 x 356/365 = 0.883 or 88.3%
However there is a formula you can apply if you studied permutation
365P10/365^10 you will get the same probability 0.883 or 88.3%

2) What is the probability that at least ten students in a class share their birthday?
The negation (or COMPLEMENT in term of probability) of "at least 10" is none of the 10
In short the complement of NON E have same date is the complement of AT LEAST 10 have the same date.
We know that P(A) + P(complement of A) =1
P(NOT same date) + P(same date ) =1
0.883 + P(same date ) =1
& P(at leas...) =0.117 or 11.7%

Answer

1 number

explanation

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

 

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.