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

[BTGmoderatorDC]

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

OA E

Source: Princeton Review

[Jay@ManhattanReview]

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

OA E

Source: Princeton Review

Given that there are 8 blue, 6 green and 4 magenta color shirts, we have 18 shirts, out of which 8 are blue and 10 are non-blue.

Probability at least one of the shirts is blue = Probability of drawing (One shirt is blue and the other is non-blue + Both shirts are blue)
= (8C1*10C1)/18C2 + 8C2/18C2 = (8C1*10C1+ 8C2)/18C2 = {8*10 + (8*7)/(1.2)} / {(18.17)/(1.2)} = 12/17

The correct answer: E

Hope this helps!

-Jay
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[Scott@TargetTestPrep]

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

OA E

Source: Princeton Review

We can use the equation:

P(at least one of the shirts he draws will be blue) = 1 - P(no blue shirts)

P(no blue shirts) = 10/18 x 9/17 = 5/9 x 9/17 = 5/17.

So, P(at least one of the shirts he draws will be blue) = 1 - 5/17 = 12/17.

Answer: E