kona wrote:
Hi Bunuel,
Please find my doubt below.
q:In a drawer of shirts, 8 are blue, 6 are green, and 4 are red. If Miles draws two shirts at random, what is the probability that at least one of the shirts he draws is blue?
My approach:1-(zero shirts blue)
1-(10*9)/18c2
1-(10/17)=7/17.
Wrong..
Another approach:1-(10/18*9/17)
1-(5/17)=12/17
Right..
My doubts:1.When picking two shirts at random from a group of 18,is it not 18c2 for a total number of possibilities
2.Also,does "picking at random" means "picking simultaneously?"
3.Does ordering in this case matter?When should the order be considered and when it shouldnt be?Could you please throw some light on this?
Please rectify
Thank you
Regards,
Kona
Hi Kona,
in your 1st approach , it should be :
1- (10C2)/(18C2) = 1 -5/17 = 12/17
Here 10C2 defines the no. of ways in which 2 shirts can be picked from 10 (6G +4R) t-shirts.
Your numerator calcn
10 *9 makes a difference b/w
a particular pair of 1G1R and 1R1G, which are essentially the same.
So,for your calcn of numerator double counting is happening.
1.Denominator 18C2 is perfectly ok for picking two shirts at random/simultaneously
2.It means the same in the present question context
3.Order would not matter as it is the case of picking (and not arranging thereafter)
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