In a drawer of shirts 8 are blue, 6 are green and 4 are

VJesus12 wrote:In a drawer of shirts 8 are blue, 6 are green and 4 are magenta. If Mason draws 2 shirts at random, what is the probability at least one of the shirts he draws will be blue?

(A) 25/153
(B) 28/153
(C) 5/17
(D) 4/9
(E) 12/17

The OA is the E.

I am really confused here. Can anyone give me some help, please? Thanks in advance.

Hello vjesus12.

We have 18 colors.

Since we want to find the probability of at least one of the shirts he draws will be blue, we can find it as follows: $$P\left(at\ least\ 1\ blue\right)=1-P\left(no\ blue\right)$$ Now, the probability of selecting no blues is: $$P\left(no\ blue\right)=\frac{10}{18}\cdot\frac{9}{17}=\frac{5}{9}\cdot\frac{9}{17}=\frac{5}{17}.$$ Hence we get: $$P\left(at\ least\ 1\ blue\right)=1-\frac{5}{17}=\frac{12}{17}.$$ Therefore, the correct answer is the option E.