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There are 1001 red marbles and 1001 black marbles in a box. Let Ps be
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20 Oct 2021, 09:18
20 Oct 2021, 09:18
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There are 1001 red marbles and 1001 black marbles in a box. Let Ps be the probability that two marbles drawn at random from the box are the same color, and let Pd be the probability that they are different colors. What is the value of |Ps − Pd|?
(A) \(0\)
(B) \(\frac{1}{2002}\)
(C) \(\frac{1}{2001}\)
(D) \(\frac{2}{2001}\)
(E) \(\frac{1}{1000}\)
(A) \(0\)
(B) \(\frac{1}{2002}\)
(C) \(\frac{1}{2001}\)
(D) \(\frac{2}{2001}\)
(E) \(\frac{1}{1000}\)
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Re: There are 1001 red marbles and 1001 black marbles in a box. Let Ps be
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20 Oct 2021, 09:44
20 Oct 2021, 09:44
Carcass wrote:
There are 1001 red marbles and 1001 black marbles in a box. Let Ps be the probability that two marbles drawn at random from the box are the same color, and let Pd be the probability that they are different colors. What is the value of |Ps − Pd|?
(A) \(0\)
(B) \(\frac{1}{2002}\)
(C) \(\frac{1}{2001}\)
(D) \(\frac{2}{2001}\)
(E) \(\frac{1}{1000}\)
(A) \(0\)
(B) \(\frac{1}{2002}\)
(C) \(\frac{1}{2001}\)
(D) \(\frac{2}{2001}\)
(E) \(\frac{1}{1000}\)
Ps = P(both are same color) = P(1st is red AND 2nd is red OR 1st is black AND 2nd is black)
= P(1st is red) x P(2nd is red) + P(1st is black) x P(2nd is black)
= 1001/2002 x 1000/2001 + 1001/2002 x 1000/2001
= 1/2 x 1000/2001 + 1/2 x 1000/2001
= 500/2001 + 500/2001
= 1000/2001
Now recognize that Pd = P(different colors) = 1 - P(both are same color)
= 1 - 1000/2001
= 2001/2001 - 1000/2001
= 1001/2001
What is the value of |Ps − Pd|?
|Ps − Pd| = |1000/2001 − 1001/2001|
= |−1/2001|
= 1/2001
Answer: C
gmatclubot
Re: There are 1001 red marbles and 1001 black marbles in a box. Let Ps be [#permalink]
20 Oct 2021, 09:44
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