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The retirement plan for a company allows employees to invest

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The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the 10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is the probability that 3 of Sam’s 4 funds grew by at least 10% over last year?

(A) 6C3 / 10C4
(B) 6C3 * 4C1 / 10C4
(C) 6C3 * 4C1 / 10P4
(D) 6P3 * 4P1 / 10C4
(E) 6P3 * 4P1 / 10P4
[Reveal] Spoiler: OA

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carcass wrote:
The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the 10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is the probability that 3 of Sam’s 4 funds grew by at least 10% over last year?

(A) 6C3 / 10C4
(B) 6C3 * 4C1 / 10C4
(C) 6C3 * 4C1 / 10P4
(D) 6P3 * 4P1 / 10C4
(E) 6P3 * 4P1 / 10P4

OA later and I would to know how you attach this problem and if is simple or quite difficult.

Thanks GMATClub


\(\frac{C^3_6*C^1_4}{C^4_{10}}\), where \(C^3_6\) is ways to select 3 funds which grew by at least 10% over the last year out of 6 such funds, \(C^1_4\) is ways to select 1 other fund from 4 and \(C^4_{10}\) is total ways to select 4 funds out of 10.

Answer: B.
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New post 21 Feb 2012, 09:36
Total funds = 10
>10% = 6
<10% = 4

Question: Prob to pick 4 funds (3>10% and 1> 10%)

Option 1: Pick 3 out of 6 = 6C3 = 20
Option 2: Pick 1 out of 4 = 4C1 = 4

Total # = 4C10 = 210

Probability = \(\frac{Option 1*Option 2}{Total}\)
or "B"

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New post 22 May 2013, 06:50
The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the 10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is the probability that 3 of Sam's 4 funds grew by at least 10% over last year?

(A) \(\frac{C^6_3}{C^10_4}\)
(B) \(\frac{C^6_3 * C^4_1}{C^(10)_4}\)
(C)
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New post 22 May 2013, 06:51
closed271 wrote:
The retirement plan for a company allows employees to invest in 10 different mutual funds. Six of the 10 funds grew by at least 10% over the last year. If Sam randomly selected 4 of the 10 funds, what is the probability that 3 of Sam's 4 funds grew by at least 10% over last year?

(A) \(\frac{C^6_3}{C^10_4}\)
(B) \(\frac{C^6_3 * C^4_1}{C^(10)_4}\)
(C)


Merging similar topics. Please refer to the solutions above.
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New post 22 May 2013, 09:39
I know this is a somewhat silly question but I need a confirmation on that

As for options (C) and (D), is it possible to combine a combination and permutation in one question? At first sight I would exclude them because I guess, it would not be logical to solve for a comb and perm. at the same time.

If this is not the case, would you please give examples of when would this be possible?
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New post 23 Feb 2016, 15:00
No math necessary:

Look at the answers. B is the only answer choice that uses only combination formulas and takes into account all the factors.

We know this is a combination, not a permutation. This rules out choices C, D & E.

A clearly isn't the answer because 6C3 only accounts for 3 of the funds he chose. 10C4, the denominator, is correct, so the numerator must be wrong.

B it is.

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New post 28 May 2016, 11:49
say that the problem stated "what is the probability that at least 3 of Sam’s 4 funds grew by at least 10% over last year?" Would we have Answer b + 6C4/ 10C4?
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New post 07 Jun 2016, 02:32
sony1000 wrote:
say that the problem stated "what is the probability that at least 3 of Sam’s 4 funds grew by at least 10% over last year?" Would we have Answer b + 6C4/ 10C4?


I am not sure and would welcome someone's professional opinion. Anyway, my guess would be:

At least 3, means 1 - P(1 or 2)
P(only 1) = 6C1 * 4C3 / 10C4 = 6*4 / 10*3*7= 24 / 210 = 12/105
P(only 2) = 6C2 * 4C2 / 10C4 = 3*5 / 10*3*7 = 15 / 210= 3/42

1 - P(only 1) + P(only 2) etc.
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