# Exploring Probability: Are You a Chocolate Lover Who Enjoys Combinations?

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Understanding probability can be as delightful as indulging in chocolates. Let's explore this with a simple example involving sweets!

Probability quantifies the chance of a specific event occurring. Recall those school lessons about determining the likelihood of picking a chocolate with a soft center from a box? If the assortment contains 7 soft centers, 5 chewy centers, and 10 solid centers, what is the probability that a randomly selected chocolate will have a soft center?

- The likelihood of selecting a soft-center chocolate is 7/22.
- The chance of picking a chewy center is 5/22.
- The probability for a solid center is 10/22.

Personally, I prefer to avoid chewy chocolates. Fortunately, these numbers make it easy to compare my chances of not getting a chewy chocolate, which is 17/22.

**Let’s get serious!** Imagine a box with 5 hard-center chocolates and 4 soft-center chocolates. Instead of devouring them all, ask someone, "**Would you like a soft-center chocolate too?**"

Secretly, I wish they wouldn’t choose a soft center, as they are my favorite.

Now, what’s the chance that both of us will randomly select a soft-filled chocolate?

We can apply the multiplication principle here. When two events occur, with x ways for one event and y ways for the other, we multiply the probabilities of both events to find the chance of both happening.

A **combination** refers to a mathematical method for determining the number of ways to arrange items where the order of selection does not matter. In **pure combinations**, items can be chosen in any order.

**First, let’s examine the combinations for choosing 2 soft-center chocolates from 9 total chocolates.**

The chance I pick a soft-center chocolate = 4/9 (as there are 4 soft centers among 9 chocolates).

The chance my friend picks a soft center = 3/8, since only 3 soft-center chocolates remain after one has been picked.

*This reflects that there are now only 3 soft-center chocolates out of the 8 remaining.*

Now, we multiply the probabilities of both events:

The probability that both of us select a soft-center chocolate is:
4/9 x 3/8 = **12/72**.

This simplifies to 2/12 and then to 1/6, or approximately one-sixth. The reason is that 6 fits into 12 **twice** and into 72 **twelve** times.

When expressed as a percentage, this results in about 17%.

This calculation employs the **General Multiplication Rule**, as detailed in the link below:
https://www.statisticshowto.com/multiplication-rule-probability

Diving deeper into combinations and probability can help us compute the chance of both of us selecting a soft-center chocolate.

Now, let’s explore the advanced realms of combinations, factorials, and permutations!

Combinatorics is a branch of mathematics focused on counting. It utilizes combinations, permutations, and factorials to solve problems. Combinations and permutations are specific cases of fundamental counting processes that simplify calculations.

Consider a box containing 9 chocolates: **5 with hard centers and 4 with soft centers.**

When drawing two chocolates, whether simultaneously or sequentially, the choice of the second chocolate depends on the first.

**Let’s derive the formula for selecting 2 chocolates with soft centers.**

To find relative frequencies (and thus probabilities) of specific events, divide the number of favorable outcomes by the total possible outcomes.

We need to assess the **COMBINATIONS** of selecting **2 soft-center chocolates from the 4 soft-center chocolates** available.

Imagine each of the 4 soft-center chocolates has a different decoration, so when you glance into the box, you see **distinct chocolates**.

Let’s visualize our soft-center treats adorned with a smiley face ?, a leaf ?, a cat's face ?, and a dog’s face ?.

*All chocolates have unique decorations, and you can’t tell which are soft or hard centers.*

In this scenario, the order in which chocolates are chosen doesn’t matter. Hence, my selection of a soft-center chocolate with a smiley face and my friend's choice of one with a leaf is equivalent to the reverse.

However, **permutations** involve arrangements where order matters.

**Returning to our example, we are calculating the probability of selecting two soft-center chocolates.**

To solve the entire problem, we need the total **NUMBER of POSSIBLE outcomes** for selecting 2 soft-center chocolates, not just from the 4 available, but from the total of 9.

The **total number of possible outcomes** is the number of ways to select 2 chocolates from 9.

Let’s compute the **combinations** of selecting 2 soft-center chocolates from the 4 available:

C(4,2) denotes the combinations of choosing 2 items from 4.

Here, n represents the total items to choose from, and r is the sample size or number of items to select.

**Using the formula with n = 4 and r = 2:**
4! / 2! x (4–2)!

The symbol "!" signifies **factorial** in mathematics, prompting you to multiply the number down until reaching one.

For example, calculating 4! (read as "four factorial"): 4x3x2x1 = 24.

Now for 2! (read as "two factorial"): 2x1 = 2.

In our formula, we also compute 4–2, which equals 2, so we use the 2! value.

**The final equation is:**
4 x 3 x 2 x 1 / [ 2 x 1 x ( 2 x 1) ].

This equals: 24 / (2 x 2) or 24 / 4 = 6.

This indicates there are 6 **combinations** for selecting 2 different soft-center chocolates from 4. Label the chocolates A, B, C, and D, and let's tabulate them.

The table illustrates 6 **combinations** of 2 soft-center chocolates from a total of 4 labeled A, B, C, and D.

This confirms a 1/6 probability of randomly selecting 2 soft-center chocolates from a total of 4.

**But wait, we also need to find the probability of choosing 2 soft-center chocolates from a mix of 9 chocolates.**

Remember, there are 4 soft centers and 5 hard centers.

We previously examined the combinations for 2 soft-center chocolates from 4.

Now, we need to analyze the combinations for selecting 2 chocolates from the entire collection of 9.

*This is because the probability of 2 soft-center chocolates is the RATIO of the NUMBER OF COMBINATIONS (or chances) of 2 soft-center treats from the 4 available, to the number of combinations (or chances) of any 2 chocolates from the total of 9.*

An **online Combinations Calculator** can help you find these results easily, but we’ll work through the problem manually for illustration.

C (9,2) denotes the combinations of 2 items from 9.

The formula for the total possible outcomes is: 9! / 2! x (9–2)!

**I love Factorials! Do you?**

Calculating 9 factorial (9!): 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880.

Calculating 2 factorial (2!): 2 x 1 = 2.

Calculating 7 factorial (7!): 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040.

Now we compute 2! times 7! as 2 x 5040, resulting in 10080.

So, 362880 divided by 10080 equals 36.

This means there are 36 combinations of 2 chocolates from 9. Let’s illustrate this visually.

4 soft centers: A, B, C, D. 5 hard centers: E, F, G, H, I. Here, highlighted in blue, are the 36 combinations, e.g., A with B, A with C, and so forth.

Next, we’ll highlight the 6 combinations of soft-center chocolates for a clearer view!

The combinations of 2 soft-center chocolates from the 4 available (A, B, C, D) include: 1. A + B 2. A + C 3. A + D 4. B + C 5. B + D 6. C + D

The probability of selecting one of these 6 from all the possibilities is calculated using the combinations formula: C(4,2) / C(9,2), meaning "divided by."

Thus, we find the ratio of the number of combinations of 2 soft-center chocolates from 4 soft-center chocolates to the number of combinations of 2 chocolates from 9 chocolates.

C(4,2) = 6, and C(9,2) = 36.

So, C(4,2) / C(9,2) gives us 6 / 36 = 1/6 (approximately 16%).

This demonstrates that there are 6 combinations of soft-center chocolates among 36 combinations of 2 chocolates from 9.

At this point, you might be exclaiming, "**Why bother with factorials when I could just calculate this as you did at the beginning?**"

Recall:
The chance that both of us select a soft-center chocolate is:
4/9 x 3/8 = **12/72**.

This simplifies to 2/12 and then to 1/6.

Using factorials is often easier for complex probability scenarios, like selecting 2 soft or hard chocolates from a mix of both.

Furthermore, the **Combinations formula** can save you from the need to list or tabulate combinations.

Understanding combinations and factorials equips you to tackle complex probability calculations confidently.

What if you wish to determine the probability of selecting 2 hard-center chocolates from the box?

**You can simplify the math as follows:**
- The chance I select a hard-center chocolate = 5/9 (5 hard centers from 9 chocolates).
- The chance my friend picks a second hard-center chocolate = 4/8.

*This results because only 4 hard-center chocolates remain after one has already been selected.*

The probability that both of us select hard-center chocolates is:
5/9 x 4/8 = **20/72**.

Calculating gives us 5 x 4 = 20 and 9 x 8 = 72.

This reduces to 10/36 or approximately 28%.

Using the table above, you can count the combinations of picking 2 hard-center chocolates.

Recall that the hard-center chocolates are labeled E, F, G, H, and I, and there are 10 combinations of 2 hard-center chocolates highlighted in green below.

This visual representation shows 10 combinations of hard-center chocolates highlighted in orange.

There are 10 combinations: 1. E + F 2. E + G 3. E + H 4. E + I 5. F + G 6. F + H 7. F + I 8. G + H 9. G + I 10. H + I

What is the probability of selecting **2 hard-center chocolates** from a mix of 4 soft-center and 5 hard-center chocolates, totaling 9?

To calculate this, we use the **Combinations formula**:
C(5,2) / C(9,2).

The number of combinations of 2 from 5 is expressed as: C(5,2) = 5! / (2! (5–2)!), simplifying to 5 / (2! x 3!).

Calculating gives us: 5 x 4 x 3 x 2 x 1 = 120, and 2! x 3! = 12.

Thus, 120 divided by 12 equals 10.

Previously, we calculated that C(9,2) is 36.

The equation for the probability of selecting 2 hard-center chocolates from 5, from a total of 9, is the ratio of: C(5,2) / C(9,2).

This results in 10 / 36 or approximately 28%.

**Now, what about the chances of getting one hard-center and one soft-center chocolate?**

The soft-center chocolates are labeled A, B, C, D, while the hard-center chocolates are labeled E, F, G, H, and I. Referring to our table, we can count **20 combinations** of one soft-center and one hard-center chocolate.

This represents the ratio of the frequency or outcomes of selecting one of each type to the total number of outcomes for selecting any 2 chocolates from 9.

To determine these frequencies, we find the combinations of 1 soft-center chocolate from 4 and 1 hard-center chocolate from 5, then multiply them, followed by calculating the ratio to the total outcomes of selecting any 2 chocolates from 9.

**The formula is:** C(4,1) x C(5,1) / C(9,2).

Calculating gives: C(4,1) = 4, and C(5,1) = 5.

C(4,1) is computed as: 4! / (1!(4–1)!) = 4 x 3 x 2 x 1 / (1 x 6) = 24 / 6 = 4.

*Logically, there are 4 ways to choose 1 soft-center chocolate from 4, and 5 ways to choose 1 hard-center chocolate from 5.*

You might wonder why we **multiply** the ways of selecting a hard-center chocolate by the ways of selecting a soft-center chocolate instead of adding them.

**Remember:**
When two events occur with x ways for one and y ways for the other, we multiply the frequencies (or ways) of both to find the chance of both occurring (i.e., one soft-center and one hard-center chocolate being chosen).

Think of it as 4 times 5 or 5 times 4 when viewing the illustration!

There are 5 ways to select one hard-center chocolate from 5 and 4 ways to select one soft-center chocolate from 4.

Thus, there are 20 combinations of one hard-center and one soft-center chocolate.

Our formula is: **C(4,1) x C(5,1) / C(9,2)**.

Calculating gives us: 5 x 4 / 36 = 20 / 36 = 0.56.

This equates to 56%, which aligns with our previously calculated values.

The probability of obtaining 2 soft-center chocolates is 16%, while for 2 hard-center chocolates, it is 28%.

If we select 2 chocolates and do not get either type, this implies we must have gotten one of each. Therefore, the probability is the remainder from 100%, calculated as follows: 16 + 28 = 44, and 100–44 = 56.

Thus, the probability of obtaining one soft-center and one hard-center chocolate is **56%**.

**Using simple math (multiplication without combinations and factorials) we could solve this as follows.**

Assuming a total of 9 chocolates, with 4 soft centers and 5 hard centers, we randomly select 2 chocolates.

**What is the probability of choosing one hard and one soft chocolate?**

**Event A:** The probability of the first selected chocolate being soft is 4 / 9.
**Event B:** The probability of the second selected chocolate being hard is 5 / 8.

The overall probability of both events occurring is 4/9 x 5/8. This is the **general multiplication rule**.

Next, consider the other possibilities:
**Event C:** The probability of the first selected chocolate being hard is 5 / 9.
**Event D:** The probability of the second selected chocolate being soft is 4 / 8.

Again, the probability of both events occurring is 5/9 x 4/8, adhering to the **general multiplication rule**.

The probability of obtaining one soft-center and one hard-center chocolate is the sum of the probabilities of the two scenarios.

It can be expressed as: (4 / 9 x 5 / 8) + (5 / 9 x 4 / 8).

This results in: 20/72 + 20/72 = 0.2777 + 0.2777 = 0.56 or 56%.

(And indeed, 56% is what we already calculated using combinations.)

There’s a 28% chance of the first chocolate being hard and the second soft, and a similar chance for the reverse.

If the two chocolates are drawn simultaneously, we can assert that there are two “mutually exclusive” **possibilities**:

- Soft-center chocolate first, hard-center second.
- Hard-center chocolate first, soft-center second.

Since order does not matter, both selections must be accounted for statistically.

See: https://math.stackexchange.com/questions/2467991/probability-of-simultaneously-selecting-multiple-objects-from-a-set https://math.stackexchange.com/questions/184115/in-need-of-tips-suggestions-when-to-add-or-multiply-probabilities https://math.stackexchange.com/questions/2195876/picking-objects-simultaneously/2195942

**Isn’t mathematics fascinating?**

Feel free to revisit the beginning of this article with boxes of hard, soft, and chewy chocolates to tackle some probability problems! ?

Combinations and permutations are crucial components of numerous probability problems, especially when the population size (n) is significant.

**Moreover, calculating combinations can prove beneficial in real life!** Many situations may arise where understanding the number of combinations is useful.

The following webpage discusses instances when you might apply combinations to determine a probability (e.g., the well-known probability of winning a lottery). The **Example 1** regarding digital PINs may have a data display issue, but you should be able to comprehend it (or work it out).

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<h2>Combination Formula: Definition, Uses in Probability, Examples & More</h2>

<div><h3>The combination formula is used to calculate the number of ways of selecting events from a collection of events, such…</h3></div>

<div><p>studyqueries.com</p></div>

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The **Example 1** also utilizes the **Permutations formula**, where the order of numbers in a digital PIN matters. A handy tip is that the Permutations formula closely resembles the Combinations formula—just without the r! in the denominator, as shown in the webpage below.

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<h2>Combinations and Permutations</h2>

<div><h3>In English, we use the term "combination" loosely, without considering if the order of items is significant. In other words…</h3></div>

<div><p>www.mathsisfun.com</p></div>

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The likelihood that you've read this far is probably slim, but if you have gained something from this article, please give it a round of applause! Thank you.

**References**
- Combinations Calculator (nCr)
- http://statstutor.ac.uk/resources/steps-glossary/glossary/probability.html
- https://courses.lumenlearning.com/boundless-statistics/chapter/probability-rules
- https://mathbitsnotebook.com/Geometry/Probability/PBIndependentDependent.html#:~:text=With%20Replacement%3A%20the%20events%20are,Probabilities%20DO%20affect%20one%20another
- https://www.onlinemathlearning.com/probability-without-replacement.html
- Permutations and Combinations
- http://onlinestatbook.com/2/probability/permutations.html
- https://readmedium.com/can-you-solve-this-intro-probability-problem-807c59543c32
- https://www.youtube.com/watch?v=DZacSLax3aM
- https://www.universalclass.com/articles/math/statistics/probability-calculate-number-of-outcomes.htm
- Understanding Probability: How to Calculate the Number of Outcomes | UniversalClass
- Probability that 2 marbles from 3 green, 6 yellow, and 4 red are different colors
- https://sciencing.com/calculate-cumulative-probability-5212997.html
- https://stattrek.com/statistics/notation.aspx#:~:text=P(A)%20refers%20to%20the,that%20event%20B%20has%20occurred.&text=b(x%3B%20n%2C%20P,refers%20to%20negative%20binomial%20probability