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805+ (Hard)|   Probability|      
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A jar contains 8 red marbles and y white marbles. If Joan takes 2 rand 10 Sep 2018, 10:53
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This is a good and tricky question. VeritasKarishma Thanks for the posts on identical and distinct objects. They are really helpful
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A jar contains 8 red marbles and y white marbles. If Joan takes 2 rand 13 Feb 2021, 21:53
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This is how I thought about it.

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
y = 8
P(RR) = 1/2 x 7/15 < P(RW) = (1/2 x 8/15)x2

y=1
P(RR) = 8/9 x 7/8 > P(RW) = (8/9 x 1/8) x2

Insufficient.

(2) y ≥ 4
y = 4
P(RR) = 2/3 x 7/11 > P(RW) = (2/3 x 4/11)x2

P(RR) > P(RW) always

Sufficient.

D
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A jar contains 8 red marbles and y white marbles. If Joan takes 2 rand 11 Jun 2026, 23:55
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I had one doubt and got response to it, posting for community

Statement - \(P(RW)=2*\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y}\), multiplying by 2 as RW can occur in two ways RW or WR;

My doubt - Why are we multiplying by 2 when WR and RW is the same thing, there is no distinction of RW means R is 1st position in race and W is 2nd position in race or R is president and W is vice-president and hence there will be 2 cases.

Here whether we take RW or WR, it is one and the same thing thus we should not have.


My mistake / Response
You're mixing up unordered selection with ordered probability calculation.

Method 1: Ordered probability approach
Suppose Joan picks marbles one at a time.
For "one red and one white", there are two possible orders:
  • RW
  • WR
    So:
    P(RW)=P(R then W)+P(W then R)
    =(8/(8+y))·(y/(7+y))+(y/(8+y))·(8/(7+y))
    =2·(8/(8+y))·(y/(7+y))
    The factor of 2 appears because the event "one of each color" can happen in two different orders.


Method 2: Combination approach
Now forget order.
Number of ways to get 1 red and 1 white:
C(8,1)·C(y,1)=8y
Number of total pairs:
C(8+y,2)
Therefore:
P(1R,1W)=8y/C(8+y,2)
No factor of 2 appears.
Why?
Because combinations already treat RW and WR as the same pair.
Why both methods give the same answer
Take the ordered version:
2·(8/(8+y))·(y/(7+y))
=16y/[(8+y)(7+y)]
Now convert the combination version:
8y/C(8+y,2)
=8y/[(8+y)(7+y)/2]
=16y/[(8+y)(7+y)]
Exactly the same.
GMAT takeaway
When using:
  • Sequential probability → count all possible orders (RW and WR), so multiply by 2.
  • Combinations → order is already ignored, so do not multiply by 2.
  • You must choose one framework and stay consistent.
    That's why the solution's statement:
    P(RW)=2·(8/(8+y))·(y/(7+y))
    is perfectly correct even though the final event is "one red and one white" without regard to order. The multiplication by 2 is compensating for the fact that the calculation is being done using an ordered draw process.


Overall solution

GMAT takeaway: when a probability comparison asks
P(2 red) > P(1 red, 1 white)?
don't calculate full probabilities unless necessary. Compare favorable cases.
Let total marbles = 8 + y.
2 red:
P(RR)=C(8,2)/C(8+y,2)
1 of each:
P(RW)=8C1*yC1/C(8+y,2) = 8*y/ yC1/C(8+y,2)
Same denominator, so compare numerators:
C(8,2)=28
Question becomes:
28 > 8y
y < 3.5
Since y is an integer,
y ≤ 3
So the DS question is really:
"Is y ≤ 3?"
Statement (1)
y ≤ 8
Could be y = 2 → answer YES.
Could be y = 8 → answer NO.
Not sufficient.
Statement (2)
y ≥ 4
Since we need y ≤ 3 for YES, every value satisfying statement (2) gives NO.
Sufficient.
Answer: B
1-minute GMAT pattern:
When comparing probabilities from the same population:
  1. Write only the favorable-count expressions.
  2. Cancel common denominators.
  3. Convert the probability question into an algebra question.
    Here:
    C(8,2) > 8y
    is much faster than computing probabilities.


Bunuel


# of total marbles in the jar equals to \(8+y\), out of which \(R=8\) and \(W=y\)

\(P(RR)=\frac{8}{8+y}*\frac{8-1}{8+y-1}=\frac{8}{8+y}*\frac{7}{7+y}\);

\(P(RW)=2*\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y}\), multiplying by 2 as RW can occur in two ways RW or WR;

Question: is \(\frac{8}{8+y}*\frac{7}{7+y}>2\frac{8}{8+y}*\frac{y}{7+y}\)? --> is \(\frac{7}{2}>y\)? --> is \(y<3.5\)? eg 0, 1, 2, 3.

(1) y ≤ 8, not sufficient.
(2) y ≥ 4, sufficient, (\(y\) is not less than 3.5).

Answer: B.
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