A box contains 19 yellow, 29 green and 37 red jelly beans.

If 14 jelly beans are selected at random, what is the probability that 5 are yellow?

I got 19C5 * 66C5 / 85C14 = 0.0000273503, but my online hw is saying this answer's incorrect. Am I missing something?

Prob(yellow) = 19/85

prob(not yellow) = 66/85

Prob(your event)
= C(14,5) (19/85)^5 (66/85)^9
= .1146..

Yeah, I just realized my mistake, it was 66C9 for the remainder of the selections, not 66C5. Thank you though

To find the probability of selecting 5 yellow jelly beans out of 14 randomly selected beans, we need to first find the total number of possible outcomes and then determine the number of favorable outcomes.

There are a total of 85 jelly beans in the box (19 yellow + 29 green + 37 red).

To calculate the total number of possible outcomes (denominator), we need to use the concept of combinations. The formula for calculating combinations is:

nCr = n! / (r!(n - r)!)
where n is the total number of items (85 in this case) and r is the number of items selected (14 in this case).

Using the formula, we can calculate:
85C14 = 85! / (14!(85-14)!) = 6,490,031,200.

Now we need to determine the number of favorable outcomes (numerator), which is the number of ways to select 5 yellow jelly beans out of the 19 available. Again, we can use the combination formula:

19C5 = 19! / (5!(19-5)!) = 10,626.

Therefore, the probability of selecting exactly 5 yellow jelly beans out of 14 randomly selected beans is:

P(5 yellow) = favorable outcomes / total outcomes
= 10,626 / 6,490,031,200
≈ 0.00000163 or 0.000163%.

So, the probability is very low, about 0.000163%.