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

Question

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

A. 1/11
 
B. 1/9
 
C. 8/33
 
D. 3/11
 
E. 1/3

broall · Accepted Answer

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?

rencsee · Answer

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GMATinsight · Answer

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

stonecold · Answer

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..

Answer must be 8/33

What am I missing ?

Bunuel · Answer

Edited the typo in options.

Heseraj · Answer

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}

Please advice me if the reasoning needs correction.

RBarbaccia · Answer

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

ScottTargetTestPrep · Answer

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.

Answer: C

BrentGMATPrepNow · Answer

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

EMPOWERgmatRichC · Answer

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

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

bumpbot · Answer

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

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