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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: 44
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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Updated on: 20 Oct 2013, 02:32
5
7
00:00

Difficulty:

25% (medium)

Question Stats:

79% (02:12) correct 21% (02:48) wrong based on 230 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

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

Originally posted by amgelcer on 19 Oct 2013, 13:47.
Last edited by Bunuel on 20 Oct 2013, 02:32, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 53063
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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20 Oct 2013, 02:45
2
1
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).

_________________
Intern
Joined: 31 Aug 2016
Posts: 46
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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12 May 2018, 06:29
Bunuel wrote:
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).

The questions wants as to find M M _ _
Math Expert
Joined: 02 Sep 2009
Posts: 53063
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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12 May 2018, 10:58
standyonda wrote:
Bunuel wrote:
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).

The questions wants as to find M M _ _

Please pay attention to the highlighted part.
_________________
Intern
Joined: 31 Aug 2016
Posts: 46
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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13 May 2018, 00:24
Bunuel wrote:

$$P(MMFF)=\frac{4!}{(2!2!)}*\frac{3}{9}*\frac{2}{8}*\frac{6}{7}*\frac{5}{6}=\frac{5}{14}$$, 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).

Please pay attention to the highlighted part.

Thanks . I was reading that it MUST have 2 males and I was also calculating MMMM and MMMF!
Intern
Joined: 31 Aug 2016
Posts: 46
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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13 May 2018, 00:55
Bunuel wrote:

$$P(MMFF)=\frac{4!}{(2!2!)}*\frac{3}{9}*\frac{2}{8}*\frac{6}{7}*\frac{5}{6}=\frac{5}{14}$$, 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).

Please pay attention to the highlighted part.

Thanks . I was reading that it MUST have 2 males and I was also calculating MMMM and MMMF!
Director
Joined: 27 May 2012
Posts: 684
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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03 Sep 2018, 04:10
1
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?

Dear Moderator,
Found this probability question , in the combination section, Hope you will do the needful. Thank you.
_________________

- Stne

Math Expert
Joined: 02 Sep 2009
Posts: 53063
Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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03 Sep 2018, 04:18
stne wrote:
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?

Dear Moderator,
Found this probability question , in the combination section, Hope you will do the needful. Thank you.

________________
Edited. Thank you.
_________________
Re: A literature panel is to select 4 panelists from a selection   [#permalink] 03 Sep 2018, 04:18
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