Problem Solving

lheiannie07 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

Can some experts show the best solution?

OA C

P(x) = # ways x can occur/# Total possibilities;

# Total possibilities: There are 12 cans, and we're selecting 2, so we want 12C2 = 12*11/2! = 66. (We can select any of the 12 cans with our first selection and any of the 11 remaining cans with our second selection. We divide by 2! because the order doesn't matter.)

# ways we can select a can of cola = 4. # ways we can select a can of root beer = 4. # ways we can select a can of cola and one can of root beer = 4*4 = 16.

16/66 = 8/33. The answer is C

lheiannie07 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

Can some experts show the best solution?

OA C

Alternatively, we can go selection by selection. Say we pick a root beer can followed by a cola can

P(selecting root beer first) = 4/12 = 1/3
P(selecting cola after having selected a root beer) = 4/11
P(selecting root beer first) * P(selecting cola after having selected a root beer) = 1/3 * 4/11 = 4/33

Of course, we could also have selected the cola first. The math will end up being identical:
P(selecting cola first) = 4/12 = 1/3
P(selecting root beer after having selected a cola) = 4/11
P(selecting cola first) *P(selecting root beer after having selected a cola) = 1/3 * 4/11 = 4/33

Now add 'em up: 4/33 + 4/33 = 8/33. The answer is C