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A literature panel is to select 4 panelists from a selection

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

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New post Updated on: 20 Oct 2013, 02:32
5
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A
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E

Difficulty:

  25% (medium)

Question Stats:

78% (02:12) correct 22% (02:44) wrong based on 228 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.
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Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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New post 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:

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

Answer: E.

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).

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

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New post 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:

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

Answer: E.

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).

Answer: E.



The question doesn't ask about MMFF.

The questions wants as to find M M _ _
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Re: A literature panel is to select 4 panelists from a selection  [#permalink]

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New post 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:

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

Answer: E.

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).

Answer: E.



The question doesn't ask about MMFF.

The questions wants as to find M M _ _


Please pay attention to the highlighted part.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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

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

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

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New post 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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

GMAT Club Bot
Re: A literature panel is to select 4 panelists from a selection &nbs [#permalink] 03 Sep 2018, 04:18
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