Probability of Pulling At Least One of Each Jellybean Flavor
Suppose you have a bag containing 5 different flavors of jellybeans. If you randomly grab 19 jellybeans from the bag, what is the probability that you pull at least one of each flavor? In this article, we will explore this question using combinatorial methods, specifically the principle of complementary counting and the inclusion-exclusion principle.
Understanding the Problem
Let's break down the problem and apply mathematical techniques to find the solution step by step.
Total Ways to Choose 19 Jellybeans
First, we need to calculate the total number of ways to choose 19 jellybeans from a bag with 5 flavors, where repetition is allowed. This can be solved using a combinatorial formula:
[ text{Total ways} binom{n k-1}{k-1} ]where
n is the number of jellybeans to choose (19 in this case). k is the number of flavors (5 in this case).Plugging in the values, we get:
[ text{Total ways} binom{19 5 - 1}{5 - 1} binom{23}{4} ]Calculating this, we have:
[ binom{23}{4} frac{23 times 22 times 21 times 20}{4 times 3 times 2 times 1} frac{212520}{24} 8855 ]Ways to Choose Jellybeans Missing at Least One Flavor
To find the number of ways to choose jellybeans such that at least one flavor is missing, we use the principle of inclusion-exclusion. This principle helps us count the number of ways that at least one flavor is missing and then subtract it from the total.
Choosing 1 Flavor to Be Missing
If we choose 1 flavor to be missing, we are left with 4 flavors. The number of ways to choose 19 jellybeans from these 4 flavors is:
[ text{Ways with 1 missing} binom{19 4 - 1}{4 - 1} binom{22}{3} ]Calculating this, we get:
[ binom{22}{3} frac{22 times 21 times 20}{3 times 2 times 1} frac{9240}{6} 1540 ]Since there are 5 ways to choose which flavor is missing, the total for this case is:
[ 5 times 1540 7700 ]Choosing 2 Flavors to Be Missing
If we choose 2 flavors to be missing, we are left with 3 flavors. The number of ways to choose 19 jellybeans from these 3 flavors is:
[ text{Ways with 2 missing} binom{19 3 - 1}{3 - 1} binom{21}{2} ]Calculating this, we get:
[ binom{21}{2} frac{21 times 20}{2 times 1} 210 ]There are (binom{5}{2} 10) ways to choose which 2 flavors are missing, so the total for this case is:
[ 10 times 210 2100 ]Choosing 3 Flavors to Be Missing
If we choose 3 flavors to be missing, we are left with 2 flavors. The number of ways to choose 19 jellybeans from these 2 flavors is:
[ text{Ways with 3 missing} binom{19 2 - 1}{2 - 1} binom{20}{1} 20 ]There are (binom{5}{3} 10) ways to choose which 3 flavors are missing, so the total for this case is:
[ 10 times 20 200 ]Choosing 4 Flavors to Be Missing
If we choose 4 flavors to be missing, we are left with 1 flavor. The number of ways to choose 19 jellybeans from this 1 flavor is:
[ text{Ways with 4 missing} binom{19 1 - 1}{1 - 1} binom{19}{0} 1 ]There are (binom{5}{4} 5) ways to choose which 4 flavors are missing, so the total for this case is:
[ 5 times 1 5 ]Applying Inclusion-Exclusion
Now we can apply the inclusion-exclusion principle to find the total number of ways to choose jellybeans missing at least one flavor:
[ text{Total missing} 7700 - 2100 - 200 5 5615 ]Calculating the Probability
The number of ways to choose 19 jellybeans such that at least one flavor is present is:
[ text{Valid ways} text{Total ways} - text{Total missing} 8855 - 5615 3240 ]Finally, the probability of pulling at least one of each flavor is:
[ P(text{at least one of each flavor}) frac{text{Valid ways}}{text{Total ways}} frac{3240}{8855} approx 0.366 ]Thus, the probability that you pull at least one of each flavor when randomly grabbing 19 jellybeans from a bag with 5 flavors is approximately 0.366 or 36.6%.
Conclusion
This problem showcases the application of combinatorial methods and the inclusion-exclusion principle in solving real-world problems. Understanding these techniques not only helps in solving complex probability problems but also enhances analytical skills in various fields such as statistics, data science, and operations research.