I have a problem calculating the part b and c of the following problem, maybe because my professor post the problem in a very wrong way. I already calculate the first part.
The problem is, A student has 8 pairs of socks all of them different colors. he will go to his class, If drawing a match before the fourth draw (with no replacement)
a)What is the probability to go to class: It is 1-(16/16)(14/15)(12/14)=.20
b) How many different color pairs of socks would you need for the probability of not go to class be greater than .90: My answer is no pair of socks
c)How many pairs do you need if you are allowed to do up to 4 draw?
(a) Is right.
Generalising: the probability of going to class with $n\geq 2$ matched pairs of socks in the drawer is: $$p_n = 1 - \frac{(2n-2)(2n-4)}{(2n-1)(2n-2)} = \frac{3}{(2n-1)} \\ p_8 = \tfrac 1 5 = 0.2$$
(b) Your answer to this is cute, though correct. However, presumably the question assumes the student is able to draw at least three socks. Find $n$ where $p_n < 0.1$
Now can you do (c) in a similar manner?