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The retirement plan for a company allows employees to inves
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09 May 2019, 01:25
09 May 2019, 01:25
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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) \(\frac{6C_3}{10C_4}\)
(B) \(\frac{6C_3*4C_1}{10C_4}\)
(C) \(\frac{6C_3*4C_1}{10P_4}\)
(D) \(\frac{6P_3*4P_1}{10C_4}\)
(E) \(\frac{6P_3*4P_1}{10P_4}\)
(A) \(\frac{6C_3}{10C_4}\)
(B) \(\frac{6C_3*4C_1}{10C_4}\)
(C) \(\frac{6C_3*4C_1}{10P_4}\)
(D) \(\frac{6P_3*4P_1}{10C_4}\)
(E) \(\frac{6P_3*4P_1}{10P_4}\)
Re: The retirement plan for a company allows employees to inves
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23 Aug 2019, 06:23
23 Aug 2019, 06:23
1
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) \(\frac{6C_3}{10C_4}\)
(B) \(\frac{6C_3*4C_1}{10C_4}\)
(C) \(\frac{6C_3*4C_1}{10P_4}\)
(D) \(\frac{6P_3*4P_1}{10C_4}\)
(E) \(\frac{6P_3*4P_1}{10P_4}\)
(A) \(\frac{6C_3}{10C_4}\)
(B) \(\frac{6C_3*4C_1}{10C_4}\)
(C) \(\frac{6C_3*4C_1}{10P_4}\)
(D) \(\frac{6P_3*4P_1}{10C_4}\)
(E) \(\frac{6P_3*4P_1}{10P_4}\)
Group the mutual funds into 2 different categories: those that grew by at least 10% and those that did not. We choose a 3 from the group that did and 1 from the lower growth group. We had a total of 10C4 to pick from
So (6C3)(4C1)/(10C4)
B
Re: The retirement plan for a company allows employees to inves
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10 Feb 2020, 20:26
10 Feb 2020, 20:26
2
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) \(\frac{6C_3}{10C_4}\)
(B) \(\frac{6C_3*4C_1}{10C_4}\)
(C) \(\frac{6C_3*4C_1}{10P_4}\)
(D) \(\frac{6P_3*4P_1}{10C_4}\)
(E) \(\frac{6P_3*4P_1}{10P_4}\)
(A) \(\frac{6C_3}{10C_4}\)
(B) \(\frac{6C_3*4C_1}{10C_4}\)
(C) \(\frac{6C_3*4C_1}{10P_4}\)
(D) \(\frac{6P_3*4P_1}{10C_4}\)
(E) \(\frac{6P_3*4P_1}{10P_4}\)
There are 6 winning funds that grew more than 10%, and 4 losing funds that grew less than 10%.
The problem can be split into 3 sub-problems:
We have the specific case where Sam must choose 4 funds, 3 of which are winning, so the remaining fund must be losing. Let’s evaluate the number of ways this can be done. [Note: The order in which the funds are chosen is not important because whether the first 3 funds are winning and the 4th one is losing, or the first fund is losing and the last 3 are winning; only 3 of 4 funds will be winning ones. Hence, this is a combination problem.]
The problem has 2 sub-problems:
1. Sam must choose 3 of the 6 winning funds. This can be done in \(6C_3\) ways.
2. Sam must choose one losing fund (say the 4th fund). There are 10 – 6 = 4 losing funds. Hence, the 4th fund can be any one of the 4 losing funds. The
selection can be done in \(4C_3\) way. Hence, the total number of ways of choosing 3 winning funds and 1 losing one is \(6C_3\) * \(4C_1\).
3. Sam could have chosen 4 funds in \(10C_4\) ways. Hence, the probability that 3 of Sam’s 4 funds grew by at least 10% over last year is \(\frac{6C_3*4C_1}{10C_4}\)
The answer is (B)
