The probability of rain tomorrow is 0.3. What does that mean?
We usually find probabilities in one of three ways:
coin tossing
what is the probability of getting “heads” when tossing a fair coin?
An experiment is a well-defined procedure that produces a set of outcomes. For example, “roll a die”; “randomly select a card from a standard 52-card deck”; “flip a coin” and “pick any moment in time between 10am and 12 am” are experiments.
A sample space is the set of outcomes from an experiment. Thus, for “flip a coin” the sample space is {H, T}, for “roll a die” the sample space is {1, 2, 3, 4, 5, 6} and for “pick any moment in time between 10am and 12 am” the sample space is [10, 12].
An event is a subset, say A, of a sample space S. For the experiment “roll a die”, an event is “obtain a number less than 3”. Here, the event is {1, 2}.
If all the outcomes of a sample space S are equally likely and if A is an event, then the probability of A is:
\[P(A)=\frac{\#\{\text{events in A}\}}{\#\{\text{events in S}\}}\]
So, the probability of an event, say A, is the ratio of success to total.
flipping a coin what is the probability of a heads?
The total number of outcomes is 2 and the number of ways to be successful is 1. Thus, P(heads) = 1/2.
consider randomly selecting a card from a standard 52-card deck: what is the probability of getting a king?
the total number of outcomes is 52 and of these outcomes 4 would be successful. So, P(king) = 4/52.
What is the probability of a sum of 8 when rolling two fair dice?
Solution 1: Sample space is
| (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |
| (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |
| (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |
| (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |
| (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |
| (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |
There are 5 pairs that have a sum of 8, so P(sum of 10)=5/36=0.1389
Solution 2: The sum can be any number from 2 to 12, the sample space is {2,3,4,..,11,12}. There are 11 numbers in the sample space, one of them is 8, so P(sum of 10)=1/11=0.091
Which is right, and why?
Let’s do a simulation to see which answer is correct. use command “sample” to randomly pick an element from a set
args(sample) shows you the correct syntax of the "sample command
sample(1:6, 2, TRUE) picks two numbers from 1 to 6 with repetition
sum(sample(1:6, 2, TRUE)) finds their sum, just what we want
z <- rep(0, 10000) #generates a vector of length 10000
for(i in 1:10000)
z[i] <- sum(sample(1:6, 2, TRUE)) #repeats our experiment 10000 times
length(z[z==8])/10000 #finds the proportion of "8's" in z## [1] 0.1381But why is it right?
The definition above works well as long as S is finite but breaks down if S is infinite. Instead modern probability, like geometry, is built on a small set of basic rules called axioms, derived in the 1930’s by Kolmogorov. They are:
\[ \begin{aligned} &\text{Axiom 1: } 0 \le P(A) \le 1 \\ &\text{Axiom 2: } P(S) = 1\\ &\text{Axiom 3: } P( \cup_{i=1}^n A_i) = \sum_{i=1}^n P(A_i)\\ \end{aligned} \]
if \(A_1, ..., A_n\) are mutually exclusive
Complement: \(P(A) = 1 - P(A^c)\)
A fair coin is tossed 5 times. What is the probability of at least one “Heads”?
Sample Space S={(H,H,H,H,H), (H,H,H,H,T), … , (T,T,T,T,T)}
S has \(2^5 = 32\) elements
P(at least one “Heads”) =
1 - P(“No Heads”) =
1 - P({(T,T,T,T,T)}) =
1 - 1/36 = 35/36
Addition Formula: \(P(A \cup B) = P(A)+P(B)-P(A \cap B)\)
We roll two fair dice. What is the probability of a sum of 5 or 8, or highest number on either die is a 3?
Sample Space is above.
Event A = {(1,4), (2,3), (3,2), (4,1), (2,6), (3,5), (4,4), (5,3), (6,2)}, n(A) = 9
Event B = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}, n(B) = 9
Event \(A\cap B = \{(2,3), (3,2)\}\), \(n(A\cap B)=2\)
\(P(A \cup B) = P(A)+P(B)-P(A \cap B)\) =
\(9/36+9/36-2/36 = 16/36 = 4/9\)