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# Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an

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Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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13 May 2017, 13:32
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Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 8/33

D. 3/11

E. 1/3
[Reveal] Spoiler: OA

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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13 May 2017, 14:02
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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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13 May 2017, 19:21
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Bunuel wrote:
Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 2/11

D. 3/11

E. 1/3

To pick randomly 2 cans out of his refrigerator, there are $$12C2=66$$ different ways.
To pick randomly 2 cans out of his refrigerator and these 2 cans are cola and beer, there are $$4 \times 4 = 16$$ different ways.

The probability is $$\frac{16}{66}=\frac{8}{33}$$

Am I missing something?
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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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13 May 2017, 20:05
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Bunuel wrote:
Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 2/11

D. 3/11

E. 1/3

Probability = $$\frac{Favourable outcomes}{Total outcomes}$$ = 1- ($$\frac{Unfavourable outcomes}{Total outcomes}$$)

Ways to pull one can of Cola = 4C1 = 4
Ways to pull one can of Root Beer = 4C1 = 4

Favourable outcome = 4C1 * 4C1 = 4*4 = 16

Total Outcomes = total ways to pull 2 cans out of total 12 cans = 12C2 = 66

Probability = 16/66 = 8/33

Bunuel Please check the options as there seems to be some mistake
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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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14 May 2017, 13:13
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I solved it via two methods ==>
Method (1)--->
which I generally use -->
Probability of the event => 4/12 *4/11 + 4/12 *4/11 => 8/33

Since it was not mentioned in the options ==> I moved to a slightly different method -->

Method 2--->

Total cases -->
(C1,C2)(C1,C3)(C1,C4)(C1,B1)...(C1,B4)(C1,G1)...(C1,G4)=> 11 cases for each item
Total cases => 11*12
Favourable cases => 4+4+4+4+4+4+4+4=> 32 cases

P(E) => 32/11*12 => 8/33 again..

What am I missing ?

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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14 May 2017, 13:25
Bunuel wrote:
Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 8/33

D. 3/11

E. 1/3

Edited the typo in options.
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Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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14 May 2017, 14:06
We can work with slot method, probability of choosing cola times probability of choosing root times its combination (in this case two factorial)

$$\frac{4}{12} *\frac{4}{11} * 2 = \frac{8}{33}$$

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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14 May 2017, 14:26
First choice is 8/12 or 2/3 because either can be chosen first. Second choice is the latter that wasn't chosen, so 4/11.

Answer: 2/3 * 4/11 = 8/33

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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18 May 2017, 19:10
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Bunuel wrote:
Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 8/33

D. 3/11

E. 1/3

Let’s first determine the number of ways Jerome can pull 1 can of root beer and 1 can of cola from the fridge. The number of ways to pull the cola is 4C1 = 4 and the number of ways to pull the root beer is 4C1 = 4.

The total number of ways to pull 2 cans from the fridge, with no restrictions, is 12C2 = (12 x 11)/2! = 66 ways.

Thus, the probability is (4 x 4)/66 = 16/66 = 8/33.

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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04 Jan 2018, 08:26
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Bunuel wrote:
Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 8/33

D. 3/11

E. 1/3

P(one can of cola and one can of root beer) = P(1st can is cola AND 2nd can is root beer OR 1st can is root beer AND 2nd can is cola)
= P(1st can is cola AND 2nd can is root beer) + P(1st can is root beer AND 2nd can is cola)
= P(1st can is cola) x 2nd can is root beer) + P(1st can is root beer) x P(2nd can is cola)
= 4/12 x 4/11 + 4/12 x 4/11
= 4/33 + 4/33
= 8/33
= C

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

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12 Jan 2018, 12:26
Hi All,

We have 4 cans each of: cola, root beer and ginger ale - which gives us a total of 12 cans. We're asked for the probability of pulling 2 cans and getting 1 cola and one root beer. The math behind this question can be approached in a couple of different ways.

1st can = cola OR root beer = 8/12

Once we pull that first can, there will then be 11 cans remaining and we'll need to pull the OTHER type (re: if we pulled cola 1st, then we would need to pull root beer 2nd - and vice versa).

2nd CAT = the other type = 4/11

Thus, the probability = (8/12)(4/11) = 32/132 = 8/33

[Reveal] Spoiler:
C

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Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an   [#permalink] 12 Jan 2018, 12:26
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