A bag contains 7 red, 6 orange, and 9 green jellybeans. What is the probability of reaching into the bag and randomly withdrawing 11 jelly beans such that the number of red ones is 7 the number of orange ones is 1 and the number of green ones is 3?
To solve this probability problem, we need to calculate the probability of picking exactly 7 red jellybeans, 1 orange jellybean, and 3 green jellybeans out of a total of 11 jellybeans.
First, let's calculate the total number of ways to pick 11 jellybeans out of the 22 jellybeans in the bag. This can be done using the combination formula:
nCr = n! / [(n-r)! * r!]
where n is the total number of items and r is the number of items we want to select.
In this case, we have a total of 22 jellybeans to choose from, so n = 22. We want to select 11 jellybeans in total, so r = 11. Therefore:
Total ways to pick 11 jellybeans = 22C11 = 22! / [(22-11)! * 11!]
Next, let's calculate the number of ways to pick exactly 7 red jellybeans, 1 orange jellybean, and 3 green jellybeans.
The number of ways to choose 7 red jellybeans out of 7 is simply 1, as there is no other choice.
The number of ways to choose 1 orange jellybean out of 6 is 6C1 = 6.
The number of ways to choose 3 green jellybeans out of 9 is 9C3 = 9! / [(9-3)! * 3!].
Therefore, the total number of ways to pick exactly 7 red jellybeans, 1 orange jellybean, and 3 green jellybeans is:
1 * 6 * (9! / [(9-3)! * 3!]) = (9! * 6) / [(6! * 3!)]
Finally, we can calculate the probability by dividing the number of ways to pick the desired combination by the total number of ways to pick 11 jellybeans:
Probability = (Number of ways to pick the desired combination) / (Total ways to pick 11 jellybeans)
Probability = [(9! * 6) / (6! * 3!)] / [22! / [(22-11)! * 11!]]
Probability = [(9! * 6 * (11! * 11!)) / (6! * 3! * 11! * (22-11)!)] / (22!)
Simplifying the expression:
Probability = (9! * 6 * 11!) / [(6! * 3! * 11!) * (11! * 11! * (22-11)!)]
Probability = 6 / (3! * (22 - 11)!)
Now, we can plug in the values to calculate the probability:
Probability = 6 / (3! * 11!)
Probability = 6 / (6 * 11!)
Probability = 1 / 11! ≈ 2.52 × 10^(-7)
Hence, the probability of randomly withdrawing 11 jellybeans such that there are 7 red ones, 1 orange one, and 3 green ones is approximately 2.52 × 10^(-7) or 0.000000252.