Addition rule for probability
P(A or B) = P(A) + P(B) – P(A and B)
Not to double count, we are subtracting the overlap.
If events are mutually exclusive, that means that they cannot happen at the same time. In that case, P(A and B) would be zero. That’s why I prefer to use the above equation rather than separating it for mutually and not mutually exclusive events.
Multiplication rule for independent events
Case 1: Flipping a single coin
What’s the P(H)?
P(H) = 1/2 and P(T) = 1/2 (they must add up to 1)
Case 2: Flipping a coin twice
What’s the P(HH)?
We can think about the all outcomes (hence our sample space):
HH
HT
TH
TT
So there’s 4 distinct, equally likely possibilities. Or, another way to think about it is, having two possibilities on the first flip and having another two possibilities on the second flip. So we have 2*2=4 possibilities.
Only 1 of those possibilities meets our criteria. So there’s 1/4 chance of that happening.
One important point: These are independent events. What happens in the first flip does not affect what happens in the second flip.
Gambler’s Fallacy
There’s something called the gambler’s fallacy, when someone thinks if they got a bunch of heads in a row, then all of a sudden, it becomes more likely on the next flip to get tails. Of course, that is not the case! Every flip is an independent event.
So we can tell P(HH) = P(H1) * P(H2) = 1/2 * 1/2 = 1/4
Case 3: Flipping a coin three times
What is P(THT)?
Note that we are not checking 2T’s in 3 flips, we are checking 3 flips in this exact order.
We can write this as P(THT) = P(T1) * P(H2) * P(T3) = 1/2 * 1/2 * 1/2 = 1/8
We can verify this, again by writing all possible outcomes:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Case 4: ‘At least one’ Head
What if we are interested in the probability of getting at least one Head out of 3 flips?
This is when the complement rule becomes handy. Because, this is the same thing as the probability of not getting all Tails in 3 flips. In other way to write this is:
1 – P(TTT) = 1 – 1/2 * 1/2 * 1/2 = 7/8
If the question would be about the probability of at least getting one Head out of 10 flips, then we could write.
1 – P(10 tails in a row) = 1 – (1/2)10 = 1 – 1/1024 = 1023/1024 ≈ 99.9%
Case 5: Another ‘At least one’ situation
A manufacturer of processing chips knows that 2% of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective?
Step 1. Find the probability that a randomly selected chip is NOT defective.
1 – 0.02 = 0.98
Step 2. Find the probability that all 4 chips are NOT defective.
(0.98)4 = 0.922
Step 3. Find the probability that at least one of the selected chips is defective.
1 – 0.922 = 0.078 = 7.8%
Generalizing the strategy
In general we can use these strategies:
P(at least 1 success) = 1 – P(all failures)
or similarly,
P(at least 1 failure) = 1 – P(all successes)
Case 6: Free-throw probability
For a basketball player (assuming has a 75% FT percentage), what are the odds of making 10 free throws in a row?
We can visualize every single throw as below:
FT #1 |- – – – – -75% – – – – – -| – -25% – – |
FT #2 |- – – – – – – – – -|- – – – |
FT #3 |- – – – – – | – – – |
…
FT #10 |-| |
Every next throw continues from the previous 75% of success. So it is in fact, 75% * 75% * … * 75%, 10 times. Of course, for a repeated multiplication, we can take an exponent. All of these are, 75% to the 10th power: (75%)10
(0.75)10 ≈ 0.06 ≈ 6%
Case 7: Probability without equally likely events
In this situation, in order to visualize the probability, we have to (kind of) take what’s called a “frequentist approach” or think about it in terms of frequency probability.
Let’s say we have an unfair coin with P(H)=0.6 and P(T)=0.4. What is the P(H1H2)?
P(H1H2) = 0.6 * 0.6 = 0.36 = 36%
What about the P(T1H2T3)?
P(T1H2T3) = 0.4 * 0.6 * 0.4 = 0.096 = 9.6%
Case 8: Test taking
On a multiple choice test, problem 1 has 4 choices, and problem 2 has 3 choices. Each problem has only one correct answer. What is the probability of randomly guessing the correct answer on both problems?
Since these events are independent, the probability will be the product of these probabilities. So it would be 1/4 * 1/3 = 1/12.
We can visualize this too.
| i1 | i2 | i3 | c | |
| i1 | ||||
| i2 | ||||
| c | This is 1/12 |
Multiplication rule for dependent events
Sample for Dependent probability: Marble
Imagine a game where we have a bag that has 2 red, 3 green marbles inside. We will pick two marbles one at a time, and won’t put back the ones we got. (Meaning that these are not independent events)
We’ll win 1 Euro if we get 2 greens out of it. If playing this game costs 35 cents, does it make sense to actually play this game?
P(1st g ∩ 2nd g) is gonna be P(1st g) times P(2nd g) given that this first was green. Another way to write this is:
P(1st g ∩ 2nd g) = P(1st g) * P(2nd g | 1st g)
= 3/5 * 2/4 = 3/10 = 30%
If we played this game many times, over and over, on average we have a 30% chance of winning 1 Euro.
So our expected value is 30% * 1 Euro = 30 cent
Now, would you want to give someone $0.35 to get on average $0.30?
Sample for Dependent probability: Coins
We have 8 coins in a bag. 3 of them are unfair and have a 60% chance of coming up Heads. We randomly choose one coin from the bag and flip it 2 times. What’s the probability of getting 2 Heads?
For this, we can draw a decision tree to visualize it.
/ 5/8 Fair —> P(HH | Fair) = 0.5 * 0.5 = 0.25
/ then P(Fair & HH) = 5/8 * 0.25 = 0.15625
\
\ 3/8 Unfair —> P(HH | Unfair) = 0.6 * 0.6 = 0.36
then P(Unfair & HH) = 3/8 * 0.25 = 0.135
But we need probability either way. Then we have to add these two.
= 0.15625 + 0.135 = 0.29125 = 29.125%
If all coins were fair, our chance would be 25%. So having unfair coins toward getting Heads makes this a little higher.
Disclaimer: Like most of my posts, this content is intended solely for educational purposes and was created primarily for my personal reference. At times, I may rephrase original texts, and in some cases, I include materials such as graphs, equations, and datasets directly from their original sources.
I typically reference a variety of sources and update my posts whenever new or related information becomes available. For this particular post, the primary source was Khan Academy’s Statistics and Probability series.