Probability

Question

Suppose a subset of 8 different days of the year is selected at
random. What is the probability that each day is from a different month? (For
simplicity, assume a year has 12 months and 30 days in each.)

Answer 1

Under the simplifying assumption each day is equally likely to be in any of the 12 months.  

 

The first day can be in any of the 12 months with P = 1

The second day must be in a different month with P = 11/12 

The third day must be in a different month than the 1st 2 with P = 10/12 = 5/6

4th day must be different from all 3 preceding with P = 9/12 = 3/4

5th day must be different from all 4 above with P = 8/12 = 2/3

6th day must be different from all 5 above with P = 7/12

7th day must be different from all 6 above with P = 6/12 = 1/2

8th day must be different from all 7 above with P = 5/12 

 

Probability that all of these things happen = product of individual probabilities = 

 

(11*5*3*2*7*1*5) / (12*6*4*3*12*2*12) = 

 

(11*5*7*5) / (12*6*4*12*12) = 1925 / 41472 ~ 0.0464