Question

A cooler contains 5 bottles of orange juice, 6 bottles of apple juice, and 4 bottles of grape juice.
Assume that both Janice and David selected orange juice. After drinking these juices, Janice and David randomly pick two more bottles from the cooler. What is the percent probability that neither will pick grape juice?

Answer 1

looks like there are 15 bottles to pick from and you want the probability that out of 4 chosen, none the four grape, meaning they are from the other 11 choices
one way to do this is to compute
\[\frac{11}{15}\times \frac{10}{14}\times \frac{9}{13}\times \frac{8}{12}\]

Answer 2

the other way is to compute
\[\frac{\binom{11}{4}}{\binom{15}{4}}\] the numerator being the number of ways to choose 4 out of the 11 not grape, and the denominator being the number of way total

Answer 3

Wait, you forgot to remove the two juices they drank.

Answer 4

i took it to mean that none were grape

Answer 5

"Assume that both Janice and David selected orange juice. After drinking these juices, Janice and David randomly pick two more bottles from the cooler." That means you want 13, not 15, and 9, not 11.

Answer 6

if that is in fact what the problem means then you would have to work in cases
none of the first two chosen are grape
one of the first two chosen are grape
both of the first two chosen are grape

compute each of these, then add them up

i am almost certain (in fact i am certain) that you will get the same answer in any case

Answer 7

The issue is that the first two chosen were not grape, but that reduces the number in the cooler. Basically, your method was perfect but you didn't take the juices they'd already drank into account.

Answer 8

how do we know the first two chosen were not grape? (i mean if you interpret the problem the way you did)?

Answer 9

OOOH DAMN!! i should learn to read!!!

Answer 10

Wait, you had "choose 4"... The problem says they start with 15 juices but they drink two orange juices.

Answer 11

it says 'orange juice"

Answer 12

ok you are right my answer is totally wrong

Answer 13

yeah i see it now, sorry i misread the problem

Answer 14

Yeah. So that changes it to a simple question of dependent events, where the first has probability 9/13 and the second has probability 8/13. There is a way to do it with factorials, but it's harder and slower.

Answer 15

\[\frac{9}{13}\times \frac{8}{12}\times \frac{7}{11}\times \frac{6}{10}\]

Answer 16

Just the first two, but you're right it should have been 8/12.

Answer 17

\[\frac{9}{13}\times \frac{8}{12}\]

Answer 18

jeez i should learn to read

Answer 19

@Rachel98 correct answer is
\[\frac{9}{13}\times \frac{8}{12}\]

Answer 20

Okay, thanks so much