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Bunuel
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 that 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


­
Total dresses in the cupboard = 8 Blue + 6 Green + 4 Magenta = 8+6+4 = 18

Mason draws 2 dresses from the cupboard.

Probability (Atleast one blue ) = ?

= 1 - Probability ( No Blues)

= 1 - (10C2 / 18 C2)

As , we remove the blues from total. We are left with 18-8 = 10 dresses which are not blue.

= 1 - [(10*9) / (18*17)]

= 1 - (10/34)

= 12/17

Option E

Alternate method :

At least ONE Blue = ( Blue and green) + ( Blue and Magenta ) + (2 blues)

= (8C1*6C1) / (18C2) + (8C1*4C1) / (18C2) + (8C2) / (18C2)

= (48+32+28)/(9*17)

= 108/153

= 12/17

Option E
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For getting the probability of choosing atleast 1 blue, the easy method is to subtract the probability of no blue from 1
Therefore:
Atleast 1 blue= 1- prob(no blue)
= 1- (10C2/18C2)
= 1- (10×9/(18×17))
= 1- 5/17
= 12/17

Option E is correct
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