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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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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.
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