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Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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09 Nov 2007, 17:19

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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?

My answer: There are total of 36 possibilities. You can construct a 6x6 table on paper or mentally to sort of visualize the number of possibilities. There are only 2 cases: 5:3 or 3:5 that one of the cards is a 5 and the sum of 2 cards is 8. Thus, the prob. is 2/36 = 1/18

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

tnguyen707 wrote:

My answer: There are total of 36 possibilities. You can construct a 6x6 table on paper or mentally to sort of visualize the number of possibilities. There are only 2 cases: 5:3 or 3:5 that one of the cards is a 5 and the sum of 2 cards is 8. Thus, the prob. is 2/36 = 1/18

since it is said that the sum is 8, total possibilities = (2, 6) (3, 5) (4,4) (5, 3)(6,2)

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

4,4 6,2 2,6 5,3 3,5

P=2/5
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1/18 would be the probability of getting an 8 with a 5-3 combination. The question has already narrowed down to the fact that the sum of the 2 cards pulled = 8. All we have to do is to look for all the possible combinations of 8, which happen to be 5. Of these, only 2 have a 5 in them. Therefore, the Prob. = 2/5

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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05 Jun 2012, 01:40

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 .

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.

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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20 Aug 2013, 22:09

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?

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.

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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26 Aug 2014, 05:32

beckee529 wrote:

i think A) 2/5

possibilities are either (5,3) or (3,5)

picking a five first + picking a five second 1* 1/5 + 1/5 * 1 = 2/5

There is something wrong in this method as there are 6 balls in the bowl and after drawing one ball the ball is put back into the bowl .. its a replacement type of problem . So it will not be 1 *1/5 + 1/5*1

Instead it will be 1*1/6 + 1/6*1 (Replacement type ).

Correct Noob Method :

Possible outcomes:

2-6 3-5 4-4 5-3 6-2 Therefore, total number of outcomes such that the sum is 8 = 5.

Number of outcome such that atleast one card in the sum is 5 = 2 (3-5,5-3).

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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27 Apr 2015, 17:06

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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

This is a conditional probability question. The formula for conditional probability says something like -

P (B/A) = P (A and B) / P(A) [Read as Prob. of event B given event A has already occured]

A: event that sum of the numbers on the 2 cards is 8 B : event that one of the 2 cards is numbered 5

Whats P(A and B) ? It means sum is 8 and one of the cards is numbered 5 Well, that's only 2 cases - (5,3) (3,5) n(favorable outcomes) = 2 n(total outcomes) = 6C2 = 15 P(A and B) = 2/15

Substitute in formula P(B/A) = Prob. that one of the 2 cards is 5 given their sum is 8 = (2/15)/(1/3) = 2/5 Ans.

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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24 Feb 2016, 14:04

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.

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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24 Feb 2016, 14:19

Engr2012 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.

But isn't the probability of selecting (4,4) twice, the same as the probability of selecting (3,5) and (5,3)?

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

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24 Feb 2016, 14:22

secterp wrote:

Engr2012 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.

But isn't the probability of selecting (4,4) twice, the same as the probability of selecting (3,5) and (5,3)?

No. When the first card is a 4 followed by another 4 how would you know which one to flip to make the second case (similar to 3,5 or 5,3)? The combination of (3,5) can be achieved by either going for 3,5 or 5,3 but the combination of 4,4 can only be 1 way 4 followed by another 4. You can not have any other way.