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probability Q

by mv2019 » Tue Jul 12, 2016 6:10 am
A jar contains 8 red marbles and 7 blue marbles. 3 marbles are drawn from the jar at random and set aside. What is the probability that exactly 2 blue marbles are among those drawn and set aside?

A) 8/65
B) 2/15
C) 24/65
D) 8/15
E) 2/3
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by GMATGuruNY » Tue Jul 12, 2016 6:26 am
mv2019 wrote:A jar contains 8 red marbles and 7 blue marbles. 3 marbles are drawn from the jar at random and set aside. What is the probability that exactly 2 blue marbles are among those drawn and set aside?

A) 8/65
B) 2/15
C) 24/65
D) 8/15
E) 2/3
P(exactly n times) = P(one way) * total possible ways.

Let B = blue and R = red.

P(one way):
ONE way to select exactly 2 blue marbles is BBR.
P(1st marble is B) = 7/15. (15 marbles, 7 of them blue.)
P(2nd marble is B) = 6/14. (14 marbles left, 6 of them blue.)
P(3rd marble is R) = 8/13. (13 marbles left, 8 of them red.)
Since we want all of these events to happen, we multiply the fractions:
P(BBR) = 7/15 * 6/14 * 8/13 = 8/65.

Total possible ways:
BBR is only ONE WAY to get exactly 2 blue marbles.
Now we must account for ALL OF THE WAYS to get exactly 2 blue marbles.
Any arrangement of the letters BBR represents one way to get exactly 2 blue marbles and 1 red marble.
Thus, to account for ALL OF THE WAYS to get exactly 2 B's and 1 R, the result above must be multiplied by the number of ways to arrange the letters BBR.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical B's:
3!/2! = 3.

Thus, P(exactly 2 blue marbles) = 8/65 * 3 = 24/65.

The correct answer is C.

More practice:
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https://www.beatthegmat.com/probability-t227448.html
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by DavidG@VeritasPrep » Tue Jul 12, 2016 8:26 am
mv2019 wrote:A jar contains 8 red marbles and 7 blue marbles. 3 marbles are drawn from the jar at random and set aside. What is the probability that exactly 2 blue marbles are among those drawn and set aside?

A) 8/65
B) 2/15
C) 24/65
D) 8/15
E) 2/3
And if you really can't get enough of these q-types, here's yet another official example: https://www.beatthegmat.com/on-a-store-c ... 71327.html
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by mjmehta81 » Thu Jul 14, 2016 10:26 am
Hello Mike,

Please help me understand why you considered arrangement in step 2, as the question is only about selection and probability. As per my calculation, answer is A.

(7C2 x 8C1) / 15C3 = 8/65.

Please explain .

Regards,

Mrudang Mehta
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by GMATGuruNY » Thu Jul 14, 2016 11:13 am
mjmehta81 wrote:Hello Mike,

Please help me understand why you considered arrangement in step 2, as the question is only about selection and probability. As per my calculation, answer is A.

(7C2 x 8C1) / 15C3 = 8/65.

Please explain .

Regards,

Mrudang Mehta
Your line of reasoning is perfect, but the value in red is incorrect.
(7C2 * 8C1)/(15C3) = 24/65.
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by Matt@VeritasPrep » Wed Jul 20, 2016 9:04 pm
Let me give one more approach, just in case you forget the combinatorics on test day:

We could draw 0 blue marbles, 1 blue marble, 2 blue marbles, or 3 blue marbles. The sum of these four probabilities must be 1, since these are the only four possibilities. We want exactly 2 marbles, so

p(exactly 2) = 1 - p(exactly 0) - p(exactly 1) - p(exactly 3)

Two of these probabilities are easy:

p(0 blue marbles) = (8/15)*(7/14)*(6/13)

p(3 blue marbles) = (7/15)*(6/14)*(5/13)

Subtracting these from 1, we now have

p(exactly 2) = 1 - (8/65) - (5/65) - p(exactly 1)

p(exactly 2) = (4/5) - p(exactly 1)

p(exactly 2) + p(exactly 1) = (4/5)

Now here's the trick. We know that the probability of one blue marble (and two red) is SLIGHTLY higher than the probability of two blue (and one red), since we have one more red marble in the jar. So p(exactly 2) should be a *little* less than (1/2) of (4/5). Only C is at all close, so this MUST be the answer.
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by Matt@VeritasPrep » Wed Jul 20, 2016 9:05 pm
Quick addendum to that last post: that approach isn't better than actually knowing the combinatorics, of course ... but it's definitely a useful way of letting the answers help you in a pinch under test conditions, and learning how to do that well is a big part of success on test day for most students.
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