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IrinaOK
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1/10 1/5 3/10 2/5 1/2 03 Sep 2007, 00:18
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1/10
1/5
3/10
2/5
1/2



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sidbidus
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1/10 1/5 3/10 2/5 1/2 03 Sep 2007, 01:03
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Is it (E)1/2???
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ywilfred
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1/10 1/5 3/10 2/5 1/2 03 Sep 2007, 01:13
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# of ways to form groups of 3 from 6 people = 6C3 = 20
# of ways to form groups with only Tom inside = 4

P = 4/20 = 1/5
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BCC145
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1/10 1/5 3/10 2/5 1/2 03 Sep 2007, 01:40
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ywilfred
# of ways to form groups of 3 from 6 people = 6C3 = 20
# of ways to form groups with only Tom inside = 4

P = 4/20 = 1/5
Show more


ywilfred,

Isn't the number of ways to from groups with Tom inside but mary 6?

4!/(2!*2!) = 6

So 3/10 fo rme.
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ywilfred
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1/10 1/5 3/10 2/5 1/2 Updated on: 03 Sep 2007, 01:44
Originally posted by ywilfred on 03 Sep 2007, 01:43.
Last edited by ywilfred on 03 Sep 2007, 01:44, edited 1 time in total.
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BCC145
ywilfred
# of ways to form groups of 3 from 6 people = 6C3 = 20
# of ways to form groups with only Tom inside = 4

P = 4/20 = 1/5

ywilfred,

Isn't the number of ways to from groups with Tom inside but mary 6?

4!/(2!*2!) = 6

So 3/10 fo rme.
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ah darn... careless mistake. yes, it should be 4C2.
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IrinaOK
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1/10 1/5 3/10 2/5 1/2 03 Sep 2007, 01:43
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BCC145
ywilfred
# of ways to form groups of 3 from 6 people = 6C3 = 20
# of ways to form groups with only Tom inside = 4

P = 4/20 = 1/5

ywilfred,

Isn't the number of ways to from groups with Tom inside but mary 6?

4!/(2!*2!) = 6

So 3/10 fo rme.
Show more


AO is C

Can you explain please?
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BCC145
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1/10 1/5 3/10 2/5 1/2 03 Sep 2007, 02:06
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IrinaOK
BCC145
ywilfred
# of ways to form groups of 3 from 6 people = 6C3 = 20
# of ways to form groups with only Tom inside = 4

P = 4/20 = 1/5

ywilfred,

Isn't the number of ways to from groups with Tom inside but mary 6?

4!/(2!*2!) = 6

So 3/10 fo rme.

AO is C

Can you explain please?
Show more


Yes sure.

Probability = Number of Expected Events / Total Number of Outcome

Total Number of Outcome = 6! / (3! * 3!) = 20

Imagine you have a group of 3 with Tom in it. So we are looking at 2 remaining spots. Also, to fill those 2 spots, we know we cannot pick Mary. So we are picking 2 people from a group of 4:

Number of Expected Events = 4! / (2! * 2!) = 6

Probability = 3/10.
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1/10 1/5 3/10 2/5 1/2 04 Sep 2007, 11:41
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I feel the OA is wrong. it has to 2/5
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dobrey
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1/10 1/5 3/10 2/5 1/2 05 Sep 2007, 01:05
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It should be 1/2.
Tom's presence is of the probability 1. Now for the rest, Mary has to be out. Which brings down the head count to 4. So probability of sleectin 2 ppl in a lot of 4 is 2/4 ie 1/2...

So effective probability is 1x1/2=1/2.

Is it right?
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bmwhype2
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1/10 1/5 3/10 2/5 1/2 24 Oct 2007, 00:32
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IrinaOK
BCC145
IrinaOK
BCC145
ywilfred
# of ways to form groups of 3 from 6 people = 6C3 = 20
# of ways to form groups with only Tom inside = 4

P = 4/20 = 1/5

ywilfred,

Isn't the number of ways to from groups with Tom inside but mary 6?

4!/(2!*2!) = 6

So 3/10 fo rme.

AO is C

Can you explain please?

Yes sure.

Probability = Number of Expected Events / Total Number of Outcome

Total Number of Outcome = 6! / (3! * 3!) = 20

Imagine you have a group of 3 with Tom in it. So we are looking at 2 remaining spots. Also, to fill those 2 spots, we know we cannot pick Mary. So we are picking 2 people from a group of 4:

Number of Expected Events = 4! / (2! * 2!) = 6

Probability = 3/10.
Got it!
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Here is a close variant of this question from MGMAT page 74.

Kate and Amy want to be on a team. There are 6 girls in the group, and only 4 will be placed on the team. What is the probability that Kate and Amy will both be on the team?
Answer= 2/5
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linau
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1/10 1/5 3/10 2/5 1/2 24 Oct 2007, 08:32
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3/10 it is...

Everything is clare with no of possible outcomes 6!/3!*3!=20
I had trouble with no of favorable outcomes. I just wrote it down
people M T A B C D, so:
TAD
TAC
TAD
TBC
TBD
TCD = 6 , so
6/20=3/10
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1/10 1/5 3/10 2/5 1/2 24 Oct 2007, 11:41
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3/10

Explanation:

n(E) = 4C2.1C1 =6 (Mary is not included)
n(S) = 6C3 = 20 (Total number of ways in which 3 out of 6 people can be selected)

P = n(E)/n(S) = 6/20 =3/10
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1/10 1/5 3/10 2/5 1/2 24 Oct 2007, 12:03
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hey bmwhype which mgmat book is this? i didnt konw that they had a book that covers probability and combinations.
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1/10 1/5 3/10 2/5 1/2 24 Oct 2007, 12:24
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jimjohn
hey bmwhype which mgmat book is this? i didnt konw that they had a book that covers probability and combinations.
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Word Translations, i believe.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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