There are 1001 red marbles and 1001 black marbles in a box. Let Ps be

Question

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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BrentGMATPrepNow · Accepted Answer

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

chetan2u · Answer

P_s  = the probability that two marbles drawn at random from the box are the same color
The first can be any of 1001+1001 or 2002, so  \frac{2002}{2002}=1 
The second should be one of the remaining 1000 marbles as one has already been picked in the first pick. =>  \frac{1001-1}{2002-1}=\frac{1000}{2001} 
 P_s=1*\frac{1000}{2001}=\frac{1000}{2001}

P_d  = the probability that two marbles drawn at random from the box are of different color
The first can be any of 1001+1001 or 2002, so  \frac{2002}{2002}=1 
The second should be one of the 1000 marbles of the different color. =>  \frac{1001}{2002-1}=\frac{1001}{2001} 
 P_d=1*\frac{1001}{2001}=\frac{1001}{2001}

|P_s − P_d|=|\frac{1000}{2001}-\frac{1001}{2001}|=|-\frac{1}{2001}|=\frac{1}{2001}

C

Fdambro294 · Answer

Might be a longer way, but I just found the probabilities of each of the 4 scenarios:

1st: P(Red 1st and then Red 2nd) + P(Black 1st and then Black 2nd) = P(S)

P(Red on 1st draw) = 1,001 / 2,002 = 1/2

and then

P(Red after drawing Red on 1st draw) = 1,000 / 2,001

= (1/2) * (1,000 / 2,001)

since we have the same number of Red as we do Black, the P(Black and then Black) will be the same, so we just double the above:

(2) * (1/2) * (1,000 / 2,001) =

1,000 / 2,0001 = P(S)

2nd: P(drawing Red 1st and then Black 2nd) + P(drawing Black 1st and then Red 2nd) = P(D)

Probability of drawing Red 1st = (1,001 / 2,002) = (1/2)
and
Probability of drawing Black given Red has been drawn = (1,001 / 2,001)

= (1/2) * (1,001 / 2,001)

P(drawing Black 1st and then Red 2nd) will be the SAME so we can just double the above like before and we get:

(2) (1/2) (1,001 / 2,001) =

1,001 / 2,001 = P(D)

Absolute Value of the Difference of: P(S) - P(D) =

(1,001 / 2,001) - (1,000 / 2,001) =

1/2,0001
C

BlueDefender · Answer

Hmm, what does the probability of Pdifferent - Psame define?

If I understand it correctly, drawing 2 marbles would mean that they are either both red/black or 1 is a different color than the other.

so what would the leftover 1/2001 mean?

Regor60 · Answer

Ps:

Number of ways to draw 2 balls of same color:

2*(1001!/2*999!)= 1001*1000

Number of ways to draw 2 balls of any color:

2002!/2*2000!= 1001*2001

Probability Ps= (1001*1000/(1001*2001/=

1000/2001

Pd:

Ways to draw a red ball is the same as ways to draw a black ball, 1001,so

1001*1001 ways to draw 2 balls of a different color.

Number of ways to draw 2 balls of any color

1001*2001 from above

Probability Pd= 1001*1001/(1001*2001)=

1001/2001

So Ps-Pd = 1000/2001 - 1001/2001=

1/2001

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There are 1001 red marbles and 1001 black marbles in a box. Let Ps be

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There are 1001 red marbles and 1001 black marbles in a box. Let Ps be

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