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A 4-person committee is to be formed from a group of 6 Democrats and

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A 4-person committee is to be formed from a group of 6 Democrats and  [#permalink]

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New post 22 Apr 2019, 21:46
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

47% (02:28) correct 53% (03:01) wrong based on 32 sessions

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A 4-person committee is to be formed from a group of 6 Democrats and 5 Republicans. Which of the following is the closest approximation to the probability that the committee will consist of at least 3 Democrats?
A)0.11
B)0.22
C)0.35
D)0.42
E) 0.75

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Re: A 4-person committee is to be formed from a group of 6 Democrats and  [#permalink]

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New post 22 Apr 2019, 23:13

Solution


Given:
    • A 4-person committee is to be formed from a group of 6 Democrats and 5 Republicans

To find:
    • The probability (closest value) that the committee will consist of at least 3 Democrats

Approach and Working:
    • Number of ways the committee can be formed with at least 3 Democrats = 3D AND 1R OR 4D = \(^6C_3 * ^5C_1 + ^6C_4 = 20 * 5 + 15 = 115\)
    • Total number of ways a 4-member committee can be formed = \(^{11}C_4 = 330\)
    • Therefore, the probability = \(\frac{115}{330} = 0.35\)

Hence, the correct answer is option C.

Answer: C

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Re: A 4-person committee is to be formed from a group of 6 Democrats and  [#permalink]

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New post 25 Apr 2019, 11:07
Abhi077 wrote:
A 4-person committee is to be formed from a group of 6 Democrats and 5 Republicans. Which of the following is the closest approximation to the probability that the committee will consist of at least 3 Democrats?
A)0.11
B)0.22
C)0.35
D)0.42
E) 0.75


Can someone solve this using the Probability method, instead of combinations?
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Re: A 4-person committee is to be formed from a group of 6 Democrats and  [#permalink]

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New post 26 Apr 2019, 07:08
Abhi077 wrote:
A 4-person committee is to be formed from a group of 6 Democrats and 5 Republicans. Which of the following is the closest approximation to the probability that the committee will consist of at least 3 Democrats?
A)0.11
B)0.22
C)0.35
D)0.42
E) 0.75


Ok, here's how I did it.

Atleast 3 Democrats = P(3D, 1R) + P(4D)

P(3D, 1R) = \(\frac{6}{11}*\frac{5}{10}*\frac{4}{9}*\frac{5}{8}\) = \(\frac{5}{66}\)
Now, there are 4 ways to get this possibility = \(4*\frac{5}{66} = \frac{20}{66}\)

P(4D) = \(\frac{6}{11}*\frac{5}{10}*\frac{4}{9}*\frac{3}{8}\) = \(\frac{1}{22}\)
Only 1 way to get this possibility.

Total probability P = \(\frac{20}{66}+\frac{1}{22} = \frac{23}{66}\)
Now, \(\frac{22}{66}=0.33\)
Therefore, \(\frac{23}{66}\) ~= 0.35.
Answer C
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Re: A 4-person committee is to be formed from a group of 6 Democrats and   [#permalink] 26 Apr 2019, 07:08

A 4-person committee is to be formed from a group of 6 Democrats and

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