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# A literature panel is to select 4 panelists from a selection

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Intern
Joined: 14 Mar 2013
Posts: 48

Kudos [?]: 77 [3], given: 119

Location: United States
GMAT Date: 12-03-2013
WE: General Management (Retail)
A literature panel is to select 4 panelists from a selection [#permalink]

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19 Oct 2013, 13:47
3
KUDOS
2
This post was
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00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:46) correct 20% (02:14) wrong based on 133 sessions

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A literature panel is to select 4 panelists from a selection of 6 female and 3 male experts. What is the probability that the selected panelists will comprise of exactly 2 males?

(A) 4/9
(B) 2/5
(C) 3/7
(D) 1/2
(E) 5/14

-----
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[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Oct 2013, 02:32, edited 1 time in total.
Edited the question.

Kudos [?]: 77 [3], given: 119

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139638 [2], given: 12794

Re: A literature panel is to select 4 panelists from a selection [#permalink]

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20 Oct 2013, 02:45
2
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Expert's post
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amgelcer wrote:
A literature panel is to select 4 panelists from a selection of 6 female and 3 male experts. What is the probability that the selected panelists will comprise of exactly 2 males?

(A) 4/9
(B) 2/5
(C) 3/7
(D) 1/2
(E) 5/14

-----
Did you try this? How about +KUDOS for me?

We are told that there should be 2 males and 2 females.

Combination approach:

$$P=\frac{C^2_3*C^2_6}{C^4_9}=\frac{45}{126}=\frac{5}{14}$$

Direct probability approach:

$$P(MMFF)=\frac{4!}{(2!2!)}*\frac{3}{9}*\frac{2}{8}*\frac{6}{7}*\frac{5}{6}=\frac{5}{14}$$, we multiply by 4!/(2!2!) because MMFF can be selected in several ways: MMFF, MFMF, FMMF, ... (the # of permutations of 4 letters MMFF, where 2 M's and 2 F's are identical).

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Kudos [?]: 139638 [2], given: 12794

Non-Human User
Joined: 09 Sep 2013
Posts: 14213

Kudos [?]: 291 [0], given: 0

Re: A literature panel is to select 4 panelists from a selection [#permalink]

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22 Dec 2017, 00:20
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Re: A literature panel is to select 4 panelists from a selection   [#permalink] 22 Dec 2017, 00:20
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