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21 Mar 2010, 22:03
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (01:58) correct
42%
(01:56)
wrong
based on 306
sessions
History
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A 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 atleast 3 of Sam's 4 funds grew by at least 10% over the last year?
A. 6C3/10C4
B. 6C3.4C1/10C4
C. 6C3.4C1+6C4/10P4
D. 6P3.4P1/10C4
E. 6C3.4C1+6C4/10C4
A. 6C3/10C4
B. 6C3.4C1/10C4
C. 6C3.4C1+6C4/10P4
D. 6P3.4P1/10C4
E. 6C3.4C1+6C4/10C4
21 Mar 2010, 22:38
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E
1) 3 out of 6 ">=10%" funds and 1 out of 4 "<10%" funds: \(C^6_3*C^4_1\)
2) 4 out of 6 ">=10%" funds and 0 out of 4 "<10%" funds: \(C^6_4*C^4_0 = C^6_4\)
3) the total number of possible selections: \(C^{10}_4\)
4) the probability is: \(\frac{C^6_3*C^4_1+C^6_4}{C^{10}_4}\)
1) 3 out of 6 ">=10%" funds and 1 out of 4 "<10%" funds: \(C^6_3*C^4_1\)
2) 4 out of 6 ">=10%" funds and 0 out of 4 "<10%" funds: \(C^6_4*C^4_0 = C^6_4\)
3) the total number of possible selections: \(C^{10}_4\)
4) the probability is: \(\frac{C^6_3*C^4_1+C^6_4}{C^{10}_4}\)
21 Mar 2010, 22:49
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mads
No of ways sam can select 4 funds = 10C4
no of ways he can select all 4 of those funds which grew by 10% =6C4
no of ways he can sleect 4 funds out of 3 of those funds which grew by 10% = 6C3*4C1
so probablity is = (6C3*4C1+6C4)/10C4
henc E is the answer
