Dan has 50 coats, and 10 of them are blue. If he randomly selects 6, what is the probability that at least 2 are blue

Question

Is a part two to this question asking what is the probability of there being fewer than 2. So if you can help with that as well it would be greatly appreciated.

Answer 1

N = 50 coats

K = 10 blue coats

n = 6 coats selected

k = # of blue coats selected

 

I use the hypergeometric distribution as opposed to the binomial distribution because this problem requires no replacement.  The table below shows where the numbers are coming from.

 

                              Selected                       Not selected                   Total

Blue coats                  k = 2                           K - k = 8                     K = 10

Non-blue coats         n - k = 4                   N + k - n - k = 36         N - K = 40

Total                          n = 6                          N - n = 44                  N = 50

 

 

P(X=0) = (40 choose 6) / (50 choose 6) = 0.2415

 

P(X=1) = (10 choose 1)(40 choose 5) / (50 choose 6) = 0.4141

 

After that, you will need to use the Complement Rule from here.

 

P(X≥2) = 1 - P(X<2) = 1 - [P(X=0) + P(X=1)] = 1 - (0.2415 + 0.4141) = 1 - 0.6556 = 0.3444

 

The probability of selecting at least 2 blue coats is 0.3444.