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by Md.Nazrul Islam » Tue Apr 03, 2012 7:55 am
In a stack of cards ,9 cards are blue and the rest are red .2 cards are to be chosen at random from the stack without replacement ,the probability that cards chosen will both be blue is 6/11. what is the number of cards in the stack ?
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by killer1387 » Tue Apr 03, 2012 8:15 am
Md.Nazrul Islam wrote:In a stack of cards ,9 cards are blue and the rest are red .2 cards are to be chosen at random from the stack without replacement ,the probability that cards chosen will both be blue is 6/11. what is the number of cards in the stack ?
9 blue and x red
9c2/(9+x)c2=6/11
=> 12/(x+9)(x+8)=1/11
=> 11*12 =(x+9)(x+8)

x=3

hence no. of cards=9+x=9+3=12
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by Bill@VeritasPrep » Tue Apr 03, 2012 8:38 am
Let x be the total number of cards.

On the first card, our probability of getting a blue card is 9/x.

On the second card, our probability of getting a blue card is 8/(x-1).

P(blue-blue) = 9/x * 8/(x-1)

6/11 = 72/(x^2 - x) (cross multiply, but don't actually do 72*11; we can factor it out)

72*11 = 6(x^2 - x) (divide both sides by 6)

12*11 = x^2 - x (factor out x on the right side)

12*11 = x(x-1)

12 = x
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