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Probability - need expert help

by voodoo_child » Wed Apr 18, 2012 5:34 am
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?

1
2/3
1/2
2/5
1/3

If sum = 8; the following combinations are possible:
(1st card, 2nd card)
2,6
6,2
3,5
5,3
4,4
4,4
OA says - Prob = 2/5 because (4,4) = (4,4). Any reasons why? Why are we not concerned about the order of the card?

Thanks
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by aneesh.kg » Wed Apr 18, 2012 5:51 am
If the sum of the cards were 8, then these are the possible cases:
(2,6), (3,5), (4,4), (5,3), (6,2)

There are 5 such possible cases.

The favourable cases are the ones in which one of the numebr is a 5.

Among the cases listed above, the two cases - (3,5) and (5,3) - are favourable.

So, 2/5 is the answer.

Note:
Say you get a 4 on the first pick and 4 on the second pick, You can get a (4,4) just once.
Can you get the same thing again? Think about it. You cannot.

But, you can get a (3,5) as well as a (5,3)
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by Anurag@Gurome » Wed Apr 18, 2012 6:05 am
voodoo_child wrote:There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?

1
2/3
1/2
2/5
1/3

If sum = 8; the following combinations are possible:
(1st card, 2nd card)
2,6
6,2
3,5
5,3
4,4
4,4
OA says - Prob = 2/5 because (4,4) = (4,4). Any reasons why? Why are we not concerned about the order of the card?

Thanks
Possible combinations of two cards that whose total is 8.
(first card, second card)
(6, 2) (2, 6) (5, 3) (3, 5) (4, 4) - only 5 possible ways sum to be 8. One from this 5 has already happened.

From these five combinations, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5 = [spoiler]2/5[/spoiler]

The correct answer is D.
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by voodoo_child » Fri Apr 20, 2012 10:46 am
aneesh.kg wrote:If the sum of the cards were 8, then these are the possible cases:
(2,6), (3,5), (4,4), (5,3), (6,2)

There are 5 such possible cases.

The favourable cases are the ones in which one of the numebr is a 5.

Among the cases listed above, the two cases - (3,5) and (5,3) - are favourable.

So, 2/5 is the answer.

Note:
Say you get a 4 on the first pick and 4 on the second pick, You can get a (4,4) just once.
Can you get the same thing again? Think about it. You cannot.

But, you can get a (3,5) as well as a (5,3)
Ok. If I agree that we cannot get (4,4) again! But let's see this example :

Two sets are defined as follows:

A - 2,3,4,4,4
B- 0,1,2

If a number is taken from set at random and another number is taken from set at random, what is the probability that the sum of these numbers is a prime integer?

Possible primes = 2,3,5,7,11 ... (11 is out)
therefore, does it matter whether you choose (4(the first),1), (4(the second),1) or (4(the third),1)? It shouldn't. In that case, the answer would be:

Total combinations = 3*3
# of ways to get 2 - (2,0)
# of ways to get 3 - (2,1) ; (3,0)
# of ways to get 5 - (4,1) (I am considering only 1 out of three cases) ; (3,2)
Answer must be 5/9, which is incorrect

Any thoughts?
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by aneesh.kg » Fri Apr 20, 2012 11:27 am
It matters. You have three 4s in your set A.

Let's take an example of Die before answering your doubt:

Every Regular Dice should be considered as a set with {1,2,3,4,5,6} where every integer from 1 to 6 comes only once.

However, If a special Dice is manufactured with (1,2,3,4,4,4} on it, then (to give you an example) the probability that a 4 appears on it when you roll it is 3C1/6C1 = 1/2. This is much more than the probability of 1/6 that you would've got with a regular dice. And, it makes sense too. The chances of a 4 appearing must shoot up compared to a regular dice because more faces have 4 on them.

So, in the question that you presented, since you have more than one 4s, the chances of Sum = 5 must also increase. Shouldn't it?
And it does. There are three ways you can get a Sum = 5:
(first 4, 1)
(second 4, 1)
(third 4, 1)
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by voodoo_child » Fri Apr 20, 2012 3:39 pm
aneesh.kg wrote:It matters. You have three 4s in your set A.

Let's take an example of Die before answering your doubt:

Every Regular Dice should be considered as a set with {1,2,3,4,5,6} where every integer from 1 to 6 comes only once.

However, If a special Dice is manufactured with (1,2,3,4,4,4} on it, then (to give you an example) the probability that a 4 appears on it when you roll it is 3C1/6C1 = 1/2. This is much more than the probability of 1/6 that you would've got with a regular dice. And, it makes sense too. The chances of a 4 appearing must shoot up compared to a regular dice because more faces have 4 on them.

So, in the question that you presented, since you have more than one 4s, the chances of Sum = 5 must also increase. Shouldn't it?
And it does. There are three ways you can get a Sum = 5:
(first 4, 1)
(second 4, 1)
(third 4, 1)

Ok - my question is, why (4(the first), 4(the second)) doesnt matter in the original example but the order matters in the second example (4(the first),1). I am still not sure. Is it because that the probability of getting a '4' in the original example is equal. However, it's not the same in the second example?
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by Mike@Magoosh » Fri Apr 20, 2012 4:52 pm
voodoo_child wrote:OK - my question is, why (4(the first), 4(the second)) doesnt matter in the original example but the order matters in the second example (4(the first),1). I am still not sure. Is it because that the probability of getting a '4' in the original example is equal. However, it's not the same in the second example?
I'll chime in, responding to a p.m. from voodoo_child.

We say that (first = 4, second = 1) is a different even from (first = 1, second = 4), because two different entities are involved. There is one physical card with a "1" on it, and another physical card with a "4" on it. Those two different cards can occur in two orders. When you switch two different things, you get a new order. AB and BA are two different combinations of letters, two different "words" (where by "word", we mean any combination of letter).

When we draw (first = 4, second = 4), there's one one physical card, the card with a "4" on it,, selected twice. When you have the same exact identical thing twice, switching the order still leaves you with the same exact identical thing. Nothing new happens. AA is the same as AA --- either way, you have the same "word".

Does this make sense? Am I getting at the heart of the question you are asking? Or is your question something different? Please let me know if there is anything else I can clarify.

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by aneesh.kg » Fri Apr 20, 2012 7:24 pm
Please don't look at it from the 'order' or 'arrangement' point of view. You just have to go to the definition of probability to understand this which says that
Probability = (No. of favourable outcomes)/ (Total no. of outcomes possible)

And answer these questions:
(i) If a regular Dice is rolled twice, what is the probability of getting a 4 on the first throw and 4 on the second throw?
(ii) If a regular Dice is rolled twice, what is the probability of getting a 4 on one of the throws and a 4 on the other throw?
(iii) If a regular Dice is rolled twice, what is the probability of getting a 2 on the first throw and 5 on the second throw?
(iv) If a regular Dice is rolled twice, what is the probability of getting a 2 on one of the throws and a 3 on one of the throws?

Now, a special dice is manufactured with {1,2,3,4,4,4} printed on it.
Answer the same questions again.
(i) If a this dice is rolled twice, what is the probability of getting a 4 on the first throw and 4 on the second throw?
(ii) If a this dice is rolled twice, what is the probability of getting a 4 on one of the throws and a 4 on the other throw?
(iii) If a this dice is rolled twice, what is the probability of getting a 2 on the first throw and 3 on the second throw?
(iv) If a this dice is rolled twice, what is the probability of getting a 2 on one of the throws and a 3 on one of the throws?

You will understand the concept when you solve these questions. Remember: follow the definition of probability.
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by spartacus1412 » Sun Apr 22, 2012 8:21 pm
it is 2/5
Its do or die this time!
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by voodoo_child » Tue Apr 24, 2012 6:42 pm
Mike@Magoosh wrote:
voodoo_child wrote:OK - my question is, why (4(the first), 4(the second)) doesnt matter in the original example but the order matters in the second example (4(the first),1). I am still not sure. Is it because that the probability of getting a '4' in the original example is equal. However, it's not the same in the second example?
I'll chime in, responding to a p.m. from voodoo_child.

We say that (first = 4, second = 1) is a different even from (first = 1, second = 4), because two different entities are involved. There is one physical card with a "1" on it, and another physical card with a "4" on it. Those two different cards can occur in two orders. When you switch two different things, you get a new order. AB and BA are two different combinations of letters, two different "words" (where by "word", we mean any combination of letter).

When we draw (first = 4, second = 4), there's one one physical card, the card with a "4" on it,, selected twice. When you have the same exact identical thing twice, switching the order still leaves you with the same exact identical thing. Nothing new happens. AA is the same as AA --- either way, you have the same "word".

Does this make sense? Am I getting at the heart of the question you are asking? Or is your question something different? Please let me know if there is anything else I can clarify.

Mike :)
Aneesh and Mike - Thanks for your thoughtful comments.

Q#1 - Let's say that if there are six cards numbered 1,2,3,4,4,4

A card is drawn and replaced in the deck. Another card is drawn and replaced again.

What is the probability that the sum of the numbers on the two cards is 5?

My solution :
1,4 (3 options each for 3 '4's)
2,3
3,2
4,1 (3 options each for 3 '4's)

Total = 8/36 Is this correct?

Q#2 - Now what is the probability that the sum will be 8?

Is the probability = (3/6) * (3/6) = 1/4?
I did 3/6 because that's the probability that a card will be '4'.

Q#3 - What is the probability that the same numbered card is drawn twice? (this will be tricky..)

Combinations (1,1); (2,2); (3,3) and thrice-(4,4)
Probability = 1/6?

Please let me know your thoughts. I am a bit confused. May be I am thinking too much.
Thanks
Voodoo Child
Last edited by voodoo_child on Thu Apr 26, 2012 8:03 am, edited 1 time in total.
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by [email protected] » Wed Apr 25, 2012 2:32 am
It is 2/5 no doubt about it... how can you differentiate between the same card. the number card 4 is one and the same whether taken first or taken second.

it is not 4A or 4B.

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by Mike@Magoosh » Thu Apr 26, 2012 10:32 am
voodoo_child wrote:Q#1 - Let's say that if there are six cards numbered 1,2,3,4,4,4
A card is drawn and replaced in the deck. Another card is drawn and replaced again.
What is the probability that the sum of the numbers on the two cards is 5?

My solution:
1,4 (3 options each for 3 '4's)
2,3
3,2
4,1 (3 options each for 3 '4's)

Total = 8/36 Is this correct?
Perfectly correct. I would just add --- reduce the fraction from 8/36 to 2/9 --- it would undoubtedly appear among the GMAT answer choices as 2/9, not 8/36. Great work! :-)
voodoo_child wrote:Q#2 - Now what is the probability that the sum will be 8?

Is the probability = (3/6) * (3/6) = 1/4?
I did 3/6 because that's the probability that a card will be '4'.
Perfectly correct. :)
voodoo_child wrote:Q#3 - What is the probability that the same numbered card is drawn twice? (this will be tricky..)

Combinations (1,1); (2,2); (3,3) and thrice-(4,4)
Probability = 1/6?
Perfectly correct. :)
voodoo_child wrote:Please let me know your thoughts. I am a bit confused. May be I am thinking too much.
My thoughts, Voodoo Child, are that you understand combinations and probability well. Great work on all of these! Let me know if there are any other questions you would like to discuss, but I'd say, you're in pretty good shape.

Mike :-)
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