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# Sam and Angela decide to join a Dance club. The club

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Senior Manager
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Sam and Angela decide to join a Dance club. The club [#permalink]

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31 Jul 2006, 08:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1. Sam and Angela decide to join a Dance club. The club includes 7 women and 7 men (including Sam and Angela). The club decides to select a woman and a man to lead the dance group. What is the probability that Sam and Angela will NOT be selected.

The Way to do this would be 1- 1/7* 1/7... But I wanted to ask that the ORDER is not important here and hence should the probab for them to be selected be 2/49 instead of 1/49.

2. What is the probab of selecting a woman and a man from a group of 7 women and 7 men.

Would this be 7C1 * 7C1/14C2 (As order does not matter here)

Basically what is the difference between the above two questions which makes the aproach different.
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31 Jul 2006, 08:16
Here is my understanding of the problem and solution.

Let's first try to find probability of the opposite event: either Sam or Angela (or both) are selected to lead the dance.

There are 7 women (including Angela) to form a possible pair with Sam.
There are 6 men (excluding Sam, who was already counted above) to pair with Angela. So, we have 13 possible pairs including either Sam or Angela.

Total number of pairs is 7x7=49.

So, the probability is 1 - 13/49 = 36/49
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31 Jul 2006, 20:03
v1rok wrote:
Here is my understanding of the problem and solution.

Let's first try to find probability of the opposite event: either Sam or Angela (or both) are selected to lead the dance.

There are 7 women (including Angela) to form a possible pair with Sam.
There are 6 men (excluding Sam, who was already counted above) to pair with Angela. So, we have 13 possible pairs including either Sam or Angela.

Total number of pairs is 7x7=49.

So, the probability is 1 - 13/49 = 36/49

its sam and angela ---- not sam or angela - its basically the couple is not selected.

1- (1/7)(1/7) = = 48/49
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31 Jul 2006, 20:15
v1rok wrote:
Here is my understanding of the problem and solution.

Let's first try to find probability of the opposite event: either Sam or Angela (or both) are selected to lead the dance.

There are 7 women (including Angela) to form a possible pair with Sam.
There are 6 men (excluding Sam, who was already counted above) to pair with Angela. So, we have 13 possible pairs including either Sam or Angela.

Total number of pairs is 7x7=49.

So, the probability is 1 - 13/49 = 36/49

its sam and angela ---- not sam or angela - its basically the couple is not selected.

1- (1/7)(1/7) = = 48/49
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31 Jul 2006, 20:46
Sorry, but I respectufully disagree

Let S be "Sam is selected",
A be "Angela is selected"

From set theory,

Event that "Sam and Angela not selected" can be written as

not(S and A) = not(S) or not(A),

and the latter translates into English as "neither Sam nor Angela selected."

I stand by my solution.

Is there OA?
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31 Jul 2006, 20:52
Oops... Even though I provided correct justification in the post just above this one, I had solved for the wrong event in my original post at the top. Of course, the answer is 48/49. I just need to learn how to read properly...
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31 Jul 2006, 21:59
1)
Lets first find out the probability of selecting Sam and Angela
Probability of Selecting Sam = 1/7
Probability that angela is selected = 1
P1 = 1/7

Probability that angela is selected = 1/7
Probabilty the Sam is selected = 1

P2 = 1/7

P = P1*P2
P = 1/49
Probability that Sam and angela are not selected = 1 - 1/49 = 48/49

2) Probability of selecting a man = 1/7
Probabiltiy of selcting a woman = 1/7

Probability of selecting a man and a woman = 1/7 * 1/7 = 1/49
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01 Aug 2006, 04:17
jaynayak wrote:
1)
Lets first find out the probability of selecting Sam and Angela
Probability of Selecting Sam = 1/7
Probability that angela is selected = 1
P1 = 1/7

Probability that angela is selected = 1/7
Probabilty the Sam is selected = 1

P2 = 1/7

P = P1*P2
P = 1/49
Probability that Sam and angela are not selected = 1 - 1/49 = 48/49

2) Probability of selecting a man = 1/7
Probabiltiy of selcting a woman = 1/7

Probability of selecting a man and a woman = 1/7 * 1/7 = 1/49

Jaynayak...

Why is this method wrong here .... 7C1 * 7C1/14C2

The order does not matter, so can't we use the above method as we did in earlier questions...
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01 Aug 2006, 19:43
sumitsarkar82 wrote:
jaynayak wrote:
1)
Lets first find out the probability of selecting Sam and Angela
Probability of Selecting Sam = 1/7
Probability that angela is selected = 1
P1 = 1/7

Probability that angela is selected = 1/7
Probabilty the Sam is selected = 1

P2 = 1/7

P = P1*P2
P = 1/49
Probability that Sam and angela are not selected = 1 - 1/49 = 48/49

2) Probability of selecting a man = 1/7
Probabiltiy of selcting a woman = 1/7

Probability of selecting a man and a woman = 1/7 * 1/7 = 1/49

Jaynayak...

Why is this method wrong here .... 7C1 * 7C1/14C2

The order does not matter, so can't we use the above method as we did in earlier questions...

jaynayak is correct if we are talking about probability that the reqd couple is selected when male-female couples are selected.

7C1 * 7C1/14C2 assumes - two men or two women can be be selected as well -
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01 Aug 2006, 21:02
old_dream_1976 wrote:
sumitsarkar82 wrote:
jaynayak wrote:
1)
Lets first find out the probability of selecting Sam and Angela
Probability of Selecting Sam = 1/7
Probability that angela is selected = 1
P1 = 1/7

Probability that angela is selected = 1/7
Probabilty the Sam is selected = 1

P2 = 1/7

P = P1*P2
P = 1/49
Probability that Sam and angela are not selected = 1 - 1/49 = 48/49

2) Probability of selecting a man = 1/7
Probabiltiy of selcting a woman = 1/7

Probability of selecting a man and a woman = 1/7 * 1/7 = 1/49

Jaynayak...

Why is this method wrong here .... 7C1 * 7C1/14C2

The order does not matter, so can't we use the above method as we did in earlier questions...

jaynayak is correct if we are talking about probability that the reqd couple is selected when male-female couples are selected.

7C1 * 7C1/14C2 assumes - two men or two women can be be selected as well -

Thats your answer. 7C! can represent men or women. Hence not conclusive.
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03 Aug 2006, 04:47
jaynayak wrote:
old_dream_1976 wrote:
sumitsarkar82 wrote:
jaynayak wrote:
2) Probability of selecting a man = 1/7
Probabiltiy of selcting a woman = 1/7

Probability of selecting a man and a woman = 1/7 * 1/7 = 1/49

Jaynayak...

Why is this method wrong here .... 7C1 * 7C1/14C2

The order does not matter, so can't we use the above method as we did in earlier questions...

jaynayak is correct if we are talking about probability that the reqd couple is selected when male-female couples are selected.

7C1 * 7C1/14C2 assumes - two men or two women can be be selected as well -

Thats your answer. 7C! can represent men or women. Hence not conclusive.

For two men or two women shouldn't it be 7C2/14C2

Once a Man has been selected we will not have 7 options to choose from...

I am a bit confused here... Plz help

Also, what if the question said - Prob for Selecting 2 men and a woman... how would we approach it then...
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03 Aug 2006, 09:21
sumitsarkar82 wrote:
1. Sam and Angela decide to join a Dance club. The club includes 7 women and 7 men (including Sam and Angela). The club decides to select a woman and a man to lead the dance group. What is the probability that Sam and Angela will NOT be selected.

The Way to do this would be 1- 1/7* 1/7... But I wanted to ask that the ORDER is not important here and hence should the probab for them to be selected be 2/49 instead of 1/49.

2. What is the probab of selecting a woman and a man from a group of 7 women and 7 men.

Would this be 7C1 * 7C1/14C2 (As order does not matter here)

Basically what is the difference between the above two questions which makes the aproach different.

Two questions are altogether different.
In the first question you can select any male except Sam AND can select any female except Angela. Th egroup of two must have one male and one female.
Total cases = 7C1 * 7C1 = 49
How many cases when Sam and Angela are selected = 1
Prob of NOT selecting these two = 1-1/49 = 48/49

Second question is asking the probability of selecting a particular man and a particular woman out of 7 men and 7 women.
In how many ways you can select any man and any women = 7C1 * 7C1 = 49
Ways when a particular man and a particular woman are selected = 1
Prob = 1/49

If second question asks: What is the probabilty of selecting 2 people from 7 men and 7 women so that the group of 2 have 1 man and 1 woman then the answer is
Total cases = 14C2 = 91
Cases when both are men = 7C2 = 21
Cases when both are women = 7C2 = 21
Cases when one is man and one is woman = 91 -42 = 49
Prob = 49/91 or 7C1 * 7C1/14C2

Hope this helps.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Re: Probability Questions   [#permalink] 03 Aug 2006, 09:21
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