This week I finished off the unit Probability and got ~30% of the way through the following unit Counting, Permutations, and Combinations. I thought both units were interesting but found the former particularly interesting and somewhat practical for the real world. Calculating the probabilities of relatively basic events were somehow simple to work through, yet difficult to fully wrap my head around. The more I did it, however, the more it started to become clear. That said, I’m quite certain I’ve only just scratched the surface of what there is to know about probability but, after getting through this unit, I realized it’s not as impenetrable as I initially assumed it would be.

The day after I made last weeks post I realized I had left out some pertinent information on the relative compliment of two sets. As a reminder, the relative compliment is:

• Relevant Compliment (“\”)
  • Can be though of as subtraction.
  • The relevant compliment of set B in A is the set of elements in A but not in B.
  • Ex. A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6}
    • A\B = {1, 3, 5}

Here is a photo from my notes that gives more information on what the relative compliment is and how to get it:

1. The image of the circle A (which should read ‘C‘) inside the rectangle U indicates that C is a subset of U. The formula you see below that image states that the relative compliment of C (C’) equals the universe, U, (which in most cases just refers to everything in a specific domain but also literally means everything) minus the set C:
   • C’ = U – C = U \ C
2. The second step indicates that this particular universe equals Z which equals all integers. You can also see that the set C now includes the numbers {-5, 0, 7}. The numbers {42, 53} that are written down inside the universe are just examples of integers that are a part of this particular universe.
3. Step 3 simply states that the integers -5, 0, and 7 are members of set C:
   • -5 ∈ C
   • 0 ∈ C
   • -7 ∈ C
4. Step 4 indicates that the integers 42 and 53 are not a part of set C. (All integers that are not a part of set C are a part of this universe.)
5. The final step, Step 5, states “the relative compliment of set C equals the universe minus set C.
   • C’ = U – C = U \ C = {42, 53, and all numbers except -5, 0, and 7}

Week 49 started out by defining what a standard Sample Space is and what’s called a Compound Sample Space in reference to probability:

• Sample Space
  • All possible outcomes of a probabilistic event.
  • Ex. the possible outcomes of a coin flip could be either {heads, or tails}, both combined making up the sample space.
• Compound Sample Space
  • When the possible outcomes of an event can be split into categories.
  • Ex. If you were choosing between 3 ice cream flavors, vanilla, chocolate, and strawberry, and between 3 different sizes, small, medium, or large, there would be 9 possible combinations making the compound sample space equal to 9.

I then finally got back to using actual math to calculate the probabilities of relatively simple events. I was given fairly straightforward questions such as “what is the likely hood of flipping a coin to land on heads 3 times in a row?”:

• P(heads, heads, heads) = ½ * ½ * ½
  • = (1/2)^3
  • = 1/8

I learned that when calculating an event involving probability X AND (⋂) Y, you multiply their probabilities together, as long as they’re independent events and not dependent:

• Independent Events
  • Probabilistic events where the outcome of one events doesn’t affect the outcome of the other event.
  • Ex. if you were to flip a coin and roll a die, the outcome of flipping the coin would have no effect on the outcome of the die.
• Dependent Events
  • The opposite of independent events where one event DOES affect the outcome of the other event.
  • Ex. trying to pick two hearts from a standard 52 card deck without putting the first card back would be considered two dependent events because:
    • P(heart_1, heart_2) = (13/52) * (13/51 OR 12/51)

The last thing I worked on in this unit was figuring out the probability of an event occurring given that (“|”) another event had occurred. Please take for example the following hypothetical table which shows the different types of clothing items and the two colours they could be:

	Blue	Green	TOTAL
Shirt	1	1	2
Hat	1		1
Scarf		1	1
Pants	1		2
TOTAL	3	3	6

In the above table, when choosing a piece of clothing at random, suppose that event A is choosing something blue and event B is choosing a shirt.

• P(A) =
  • = 3/6
  • = ½
  • = 0.5
• P(B) =
  • = 2/6
  • = 1/3
  • = 0.333
• P(A|B) =
  • (This means, “given that you’ve chosen a shirt, i.e. B, what’s the probability it’ll be a blue, A, shirt?”)
  • = ½
  • = 0.5
• P(B|A) =
  • =1/3
  • = 0.333
• P(A⋂B) =
  • (“What’s the probability of choosing a blue, A, shirt, B?
  • = A * B
  • = 3/6 * 2/6
  • = ½ * 1/3  
  • = 0.5

When stating the probability of one event occurring ‘given that’ another event has occurred, you’re essentially limiting that sample space to the possible outcomes within the given set. It’s worth noting the following formulas tell you if events are independent or dependent:

• Independent
  • P(A) = (A|B)
  • P(B) = P(B|A)
• Dependent
  • P(A) ≠ P(A|B)
  • P(B) ≠ P(B|A)

I still don’t fully understand this concept at this point. I think it basically means, if the probability of A occurring is the same as A occurring after B has occurred, since B hasn’t altered the likelihood of A occurring, the events are independent.

The last thing I learned this week when I began the unit Counting, Permutations, and Combinations was what ‘permutations’ are exactly. Here are 2 pages from my notes to better help describe them:

As you can see, the initial question asks to find how many possible seating arrangements there are for 5 people on 5 chairs. To begin, there are 5 possible choices for the first chair, then 4 possible choices for the second chair, 3 for the third, 2 for the fourth, and 1 for the fifth. The way to calculate how many combinations are possible would be:

• Combinations = 5 * 4 * 3 * 2 * 1
  • = 15, 600

This type of math is called a permutation where you begin with any given number, n, then subtract 1 from that number, i.e. n – 1, and multiply the difference, and repeat until you’ve gone from the nth number to 0. In the above example, you can write 5 * 4 * 3 * 2 * 1 as 5! which literally translates to ‘five factorial’. In the same way multiplying two numbers gives you their product, taking a numbers factorial gives you its’ permutation (I believe, although I could be wrong about that). The formula for a factorial is:

• n! = n * (n – 1) * (n – 2) * … * 1

Lastly, if you’d like to find the value of a specific number of iterations of a factorial (i.e. the second question shown on the second page above), you divide the factorial by n – r where r equals the number of iterations you’re trying to calculate. For example, if you’d only like to find the value of the first three iterations in the above example:

• 5! / 2! =
  • = (5 * 4 * 3 * 2 * 1) / (2 * 1)
  • = (5 * 4 * 3 * 2 * 1) / (2 * 1)
  • = 5 * 4 * 3
  • = 60

Since you’re only trying to find the first three iterations, you divide the numerator by two. (I’m fairly certain I’m not explaining this all that well and without the proper terminology but this is how I understand it up to this point.)

Overall this week was semi tough but fairly enjoyable. It was nice once again working through equations as opposed to simply copying and memorizing definitions. I still find the concept of independent vs dependent variables tough to understand so hopefully I’ll get a bit more practice on those types of questions in the future. I should be able to finish this current unit Counting, Permutations, and Combinations (160/500 M.P.) within the next two days and hopefully will get through a good chunk of the following unit Random Variables (0/2100 M.P.). I’m also thinking of setting another goal of finishing all of the statistics courses by the end of October. Once I’m closer to finishing this course I’ll have a better idea of how much I’ll have remaining in the other two courses which will help me determine if finishing stats by November is a realistic goal. It’s starting to look more and more doable however since, as I go through this current course, Statistics and Probability, much of the following two units are also being completed. My New Year’s Resolution was to at least get into calculus before the New Year so, at the very least, I should accomplish that!