Check out some similar questions!

SourPatchKids · Feb 6, 2011, 01:13 AM

I just finished a bag of Sour Patch Kids. There are four different colored Sour Patch Kids in each bag, those being yellow, green, orange, and red. With seven Sour Patch Kids left in the bag, I reached in for three at once. I pulled out three greens. I reached in for three Sour Patch Kids once again from the remaining four. I pulled out three yellows. The one Sour Patch Kid remaining was a red.

What is the probability of all of these events occurring?

Unknown008 · Feb 6, 2011, 04:48 AM

The probability of taking out 1 green is 3/7 (3 greens out of total 7)
The probability of taking out a second green is 2/6 (6 greens out of total 6)
The probability of taking out a third green is 1/5 (1 green out of total 5)

The same logic is used for the other Sour Patch Kid. When you get them all, multiply each of them together, that is:

$Probability = \frac37 \times \frac26 \times \frac15 \times ...$

SourPatchKids · Feb 6, 2011, 01:13 PM

What about the fact that, out of an entire bag of Sour Patch Kids, three greens, three yellows, and one red remained (assuming that the factory selection is completely random and we have four possibilities). I want to include both "events" into the equation. I get somewhere around 40 millionths, but that seems a bit unlikely.

Unknown008 · Feb 7, 2011, 12:34 AM

Hm.. that will depend on how many there are before opening the packet.

But then, this might be a very long and tedious work... since the number of each colour is random at the production.

SourPatchKids · Feb 7, 2011, 02:50 AM

Theoretically, the first seven Sour Patch Kids put into the bag could be the last seven pulled out (a probability in and of itself, I know). Assuming that each color selected at production is random, would it make any sense to assign a 1/4 probability to each selection because of the four possibilities? And following that logic, is then raising 1/4 to the seventh power an accurate way of calculating the probability that the last seven Sour Patch Kids will be three greens, three yellows, and one red?

Unknown008 · Feb 7, 2011, 06:34 AM

Hm... point taken. That is if you don't know the exact amount and colour of the Sour Patch Kids, then it will amount to approximately (1/4)^7

It then becomes like this problem:
There is a huge basin of Sour Pack Kids, with 4 colours and the probability of picking a particular colour is equal to the probability of picking any other colour. What is the probability of picking 3 G, 3 Y and 1 R?

Here, the probability of picking 1 G tends to 1/4, another green again to 1/4, etc to (1/4)^7