Question

We consider the experiment of selecting a college student at random and looking at their second initial, i.e., the first letter in their family name. For purposes of this problem, we Date: February 12, 2020. 1 will pretend that all the outcomes are equally likely. (In practice, some letters are much more common than others.) How many students must a class have (a) To make it impossible for all the students in the class to have distinct second initials? (b) To make the probability that all the students in the class have distinct second initials less than 50%?

Answer 1

a) There are a total of 26 alphabets in english language, therefore the minimum number of students in the class that would make it impossible for all the students in the class to have distinct second initials would be 27 as there are only 26 distinct initials possible. Therefore 27 is the required answer here.

b) Let number of students needed to make the probability that all the students in the class have distinct second initials less than 50% be n. Then, we have here:

Prob. that all are distinct < 0.5

(26! / (26 - n)! ) / 26^n < 0.5

We can only do this by hit and trial method.

For n = 10, we have here LHS computed as:
(26! / 16!) / 26^10 = 0.1365

For n = 5, we have the LHS as:
(26! / 21!) / 26^5 = 0.6644

For n = 6, we have the LHS as:
(26! / 20!) / 26^6 = 0.5366

For n = 7, we have the LHS as:
(26! / 19!) / 26^7 = 0.4128

Therefore n = 7 is the required sample size here.