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# There are 6 cards numbered from 1 to 6. They are placed into a box, an

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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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I'm curious why everyone is counting 4-4 as only one possibility. You could select the first four and then the second four, just like you could have the separate scenario of selecting the second four and then the first four. They are separate cards, aren't they? These two scenarios are just as plausible as selecting 2-6 and 6-2 as different outcomes. What's wrong with my logic?
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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zebing wrote:
I'm curious why everyone is counting 4-4 as only one possibility. You could select the first four and then the second four, just like you could have the separate scenario of selecting the second four and then the first four. They are separate cards, aren't they? These two scenarios are just as plausible as selecting 2-6 and 6-2 as different outcomes. What's wrong with my logic?

(4, 4) is one case: first draw = 4 and second draw = 4.

While (2, 6) and (6, 2) are 2 different cases:
First draw = 2 and second draw = 6;
First draw = 6 and second draw = 2.

Hope it's clear.
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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Can someone explain to me why (4,4) is only counted once?
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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secterp wrote:
Can someone explain to me why (4,4) is only counted once?

Because if you end up getting 4,4 then it doesnt matter if your first card was 4 or your second card was 4. The total will still be 8. For unequal cards, it does matter whether you draw a 5 first and then a 3 or if you draw a 3 first and then a 5. These are 2 different cases.

You will end up double counting 4,4 case if you are not careful.
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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ggarr wrote:
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?

A. 2/5
B. 3/5
C. 1/18
D. 1/3
E. 1/6

1. The total number of cases is restricted by the fact that the sum of the numbers of the cards drawn be 8. So total number of cases instead of being 6*6 will be only those where the sum of the numbers is 8. The cases are (2,6), (6,2), (5,3), (3,5), (4,4)
2. The favorable outcome is in two such cases (5,3) and (3,5)
3.So the probability is 2/5
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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Hi All,

We're told that six cards numbered from 1 to 6 are placed in an empty bowl; one card is drawn at random and then put back into the bowl and then a second card is randomly drawn. We're told that the sum of the numbers on the cards is 8. We're asked for the probability that one of the two cards drawn is a 5.

This question involves some basic Probability math, but you might end up doing some unnecessary math along the way.... Here's what that unnecessary math would look like:

With 6 cards, there are (6)(6) = 36 possible pairs that can be drawn.

Of those 36 pairings, there are only a five options that actually add up to 8:
2-6
3-5
4-4
5-3
6-2

Thus, the probability of getting a sum of eight = 5/36

Of those five options, only two of them include a '5', so the probability of pulling a 3-5 or a 5-3 = 2/36

Thus, the overall probability = (2/36)/(5/36) = 2/5

It's worth noting that you can actually avoid most of these steps entirely if you focus on the specific 'pairings' that the prompt asks about. The "given" is that the sum is equal to 8, meaning that we don't actually have to calculate the probability of getting that sum. There are 5 ways to do that. Once we list them out, there clearly only 2 options (of the 5) that fit what we're looking for.... so the answer is 2/5.

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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
Bunuel wrote:
Joy111 wrote:
ggarr wrote:
Quote:
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?

2/5
3/5
1/18
1/3
1/6

My soln:

two possibilities:

5,3 and 3, 5

1/6*1/6*2= 1/18

some answers say 2/5 , but since the cards can be replaced , I think 2/5 may not be the correct answer .

can anyone shed some light please ? Thank you

What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

Check: https://gmatclub.com/forum/the-probabili ... 85523.html for more.

Hope it's clear.

Why cant answer be like 2/36. 2 being the desired no of cases and 36 being total no of cases.
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]

GMAT questions are always carefully worded - and with Quant questions, it might help to think of the details of the question as 'restrictions' (that essentially 'limit' what you should be thinking about and the type of math that you might work through).

Here, the question tells us that the sum of the two cards MUST be 8, so we have to restrict the set of possibilities to JUST that group. Based on what we are told about the cards and how they are to be drawn, there are only five options that would give us a total of 8. Since we're NOT considering anything OTHER than a total of 8, we can't be thinking about all 36 possible pairs of cards (as they're not all relevant to how the question is written). The options that ARE permissible are:

2-6
3-5
4-4
5-3
6-2

From this group of five, there are two options that 'fit' what we're looking for (3-5 and 5-3), so the answer to the question "IF... the sum on the two cards is 8, then what is the probability of one of the cards being a 5?".... is 2/5.

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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
Bunuel wrote:
Joy111 wrote:
ggarr wrote:
Quote:
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?

2/5
3/5
1/18
1/3
1/6

My soln:

two possibilities:

5,3 and 3, 5

1/6*1/6*2= 1/18

some answers say 2/5 , but since the cards can be replaced , I think 2/5 may not be the correct answer .

can anyone shed some light please ? Thank you

What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

Check: https://gmatclub.com/forum/the-probabili ... 85523.html for more.

Hope it's clear.

Why cant answer be like 2/36. 2 being the desired no of cases and 36 being total no of cases.

We are given that the sum of the numbers on the cards is 8.
This means that my acceptable universe is of only those cases in which the sum is 8.
There are only 5 such cases (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).
Hence the denominator will not be 36 but 5.
Only in 2 of these 5 cases, do we pick the card number 5.
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
ENGRTOMBA2018 wrote:
secterp wrote:
Can someone explain to me why (4,4) is only counted once?

Because if you end up getting 4,4 then it doesnt matter if your first card was 4 or your second card was 4. The total will still be 8. For unequal cards, it does matter whether you draw a 5 first and then a 3 or if you draw a 3 first and then a 5. These are 2 different cases.

You will end up double counting 4,4 case if you are not careful.

In the same way that 5, 3 & 3,5 are counted twice, 4,4 and 4,4 should be counted twice, I still do not understand the difference. Can somebody explain to me this again?
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
Hi Akunamatata,

We're not really counting an option "twice"; the prompt tells us that the cards are drawn individually (the first card, which is then put back in the bowl - then the second card). Thus, you have to think in those specific terms when you list out all of the options. They would be:

-first card 2, second card 6
-first card 3, second card 5
-first card 4, second card 4
-first card 5, second card 3
-first card 6, second card 2

That is a total of 5 options that sum to 8.

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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
Bunuel wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8?
(A) 1
(B) $$\frac{2}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{2}{5}$$
(E) $$\frac{1}{3}$$

What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

Check: https://gmatclub.com/forum/the-probabili ... 85523.html for more.

Hope it's clear.

Bunuel Requesting for your help here! This is my weakest area

Please tell me why am I wrong?

Facts:
(i) We select numbers at random
(ii) Only possible ways to get a sum of 8 in this situation are: (5,3) & (3,5)

Now. there are 2 cases possible:

Case 1: 5 is selected first, then 3
Probability = (1/6)*(1/6) = 1/36

Case 2: 3 is selected first, then 5
Probability = (1/6)*(1/6) = 1/36

Total Probability = (1/36) + (1/36) = 1/18

Why is this wrong?
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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Bunuel wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8?
(A) 1
(B) $$\frac{2}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{2}{5}$$
(E) $$\frac{1}{3}$$

What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

Check: https://gmatclub.com/forum/the-probabili ... 85523.html for more.

Hope it's clear.

Bunuel Requesting for your help here! This is my weakest area

Please tell me why am I wrong?

Facts:
(i) We select numbers at random
(ii) Only possible ways to get a sum of 8 in this situation are: (5,3) & (3,5)

Now. there are 2 cases possible:

Case 1: 5 is selected first, then 3
Probability = (1/6)*(1/6) = 1/36

Case 2: 3 is selected first, then 5
Probability = (1/6)*(1/6) = 1/36

Total Probability = (1/36) + (1/36) = 1/18

Why is this wrong?

QUESTION 1: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that the sum of the two cards will be 8?

P(sum of 8) = P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) + P(4, 4) = 1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 + 1/6*1/6 = 5/36.

QUESTION 2: There are 6 cards numbered from 1 to 6. They are placed into a box, and then two cards are drawn without replacement. What is the probability that the sum of the two cards will be 8?

P(sum of 8) = P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) = 1/6*1/5 + 1/6*1/5 + 1/6*1/5 + 1/6*1/5 = 2/15. Here (4, 4) case is not possible since we have only one 4 in the box and we are not putting back cards we draw.

ORIGINAL QUESTION: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 IF the sum of the two was 8?

This one is different from the two above because We are not asked to find the probability of getting the sum of 8, we are told that the sum of 8 has already occurred - we have the sum of 8. So, the question asks: if we have the sum of 8, what is the probability that one of the cards was 5?

Now, considering all above check the solution provided.

What combinations of two cards are possible to total 8?
(first card, second card):
(6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

Does this make sense?
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
secterp wrote:
Can someone explain to me why (4,4) is only counted once?

Imagine yourself trying to pick numbers such that there sum is 8.
If you pick the first number and see a 1 then you will be sad cause now its impossible for you to get an eight.
However if you get a 3 in your first pick then you will be praying hard to get a 5 on the second pick
On the other hand, If you get a 5 in your first pick then you would want to get a 3 in your second pick.
As you can see here there are 2 ways in which you can combine 3 and 5 in order to get to 8.
This is not the case if you want to combine 4 and 4 in order to get to 8. Why? Because in the first case when you wanted to combine a 3 and a 5 to get 8 then if you get a 3 in the first pick, its still not dead you can still achieve your goal by getting a 5, and if you get a 5 in your first pick then its still not dead, you can get a 3 in your second pick.
While if you are just relying on a 4 and 4 to get to the 8, anything other than a 4 in your first pick kills your chances in achieving your goal. And so thats why you count it as 1 case and not 2 cases.
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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To understand underlying concept better, try the following similar questions:
https://gmatclub.com/forum/there-are-6- ... 80664.html
https://gmatclub.com/forum/there-are-6- ... 80662.html
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
If the cards add to 8, that gives us these possibilities (remember that choosing (a,b) will be distinct from choosing (b,a)):

2,6 and 6,2

3,5 and 5,3

4,4

That's five different possibilities that sum to 8, of which two contain a 5.

Therefore the probability is 2/5 or answer A

Here's a video of my solution to the same problem:

https://youtu.be/qM_DyluBVP8
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Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
What a beautifully crafted Official Question of Probability! Simple, elegant, can be solved less than 30 secs applying the correct logic! WOW
Re: There are 6 cards numbered from 1 to 6. They are placed into a box, an [#permalink]
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