A = {1, 2, 3, 4, 5}
B = {}
(also Universe of Discourse or Sample Space)
U or S
A or B = {x | x is a subset of A or x is a subset of B}
A and B = {x | x is a subset of A and x is a subset of B}
Not A = {x | x is a subset of U but x is not a subset of A}
We flip a balanced coin three times (or flip three coin simultaneously). The sample space of this experiment is:
U = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let's define event A thus:
What is the probability of A?
We observe from the sample space that there are exactly three sample points that match event A (HHT, HTH, and THH). Hence, P(A) = 3/8. (The assumption, of course, is that all the sample points in the sample space are equally likely, since we are told that the coin is “balanced.”)
We toss a pair of dice simultaneously. There are 36 events (or sample points) in the sample space of this experiment:
| 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 |
Let's define events A, B, C, D thus:
What are the probabilities of A, B, C, and D? What is P(C or D)
How many ways can we permute (or rearrange) the letters ABC? In other words, in how many ways can we permute ABC (3 letters) taken 3 at a time?
Formally, the number of permutation of m items taken n at at time is given as m!÷(m-n)!. So we have:
3!/(3-3)! = 3!/0! = 3! = 3 * 2 = 6 (0! is taken to be equal).
The 6 ways are: ABC, ACB, CBA, CAB, BAC, BCA.
How many ways can we permute (or rearrange) the letters ABC, taken 2 at at time?
3!/(3-2)! = 3!/1! = 3 * 2 = 6
The 6 ways are: AB, AC, BC, BA, CA, CB.
There are 52 cards in a deck of cards. Let's define event A thus:
Find P(A) and P(B).
The number of combinations of m items taken n at at time is given as m!÷[n!(m-n)!]. So we have:
P(A) = 52! ÷ [5!(52-5)!] = 52! ÷ (5!47!) = 2,598,960.
2,598,960 represents the number of combinations of five cards that we could draw from a deck of card (52 cards).
P(B) = n(B) ÷ n(S) = [13! ÷ [5!(13-5)!]] ÷ 2598960 (that is, the number of combinations of 13 hearts taken 5 at a time over the number of combinations of five cards that could be drawn from a deck of 52 cards) =
How many different ways can we rearrange the letters in the word, RADAR.
(Hint: While RADAR could be a permutation of RADAR (as in, we switch the first and last Rs), it does not constitute of distinct way of rearranging the word, and hence, is not a distinct combination of it).
The number of distinct way of rearranging the letters of the word RADAR is given by
5! ÷ (2!2!1!) = 30
Because we are interested in distinct arrangement, we must account for the repetitions. Specifically, there are 2 occurences of Rs and As and 1 occurence of D (hence the division by 2!2!1!).
§
So then, how many different (distinct) combinations are there of MISSISSIPPI?
Our experiment is to toss a coin 3 times and observe the outcomes. The sample space of this experiement contains 8 sample points: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. So n(S) = 8. Let us define event A thus:
A: Observe exactly 2 Hs
n(A) = 3, n(S) = 8, so P(A) = 3/8 = .375
Suppose we are told that the first toss resulted in a H, and are asked to find the probability that we will observe a H on the second toss. Let's define event B thus
B: Observe a H on second toss given that a H was observed on the first
We will represent the preceding expression, called a conditional probability as P(A|B).
Because we know that an H was observed on the first toss, the sample space for event B is reduced to HHH, HHT, HTH, HTT, that is, to n(S) = 4. And of those 4 sample points, only 2 represent event B, so that
P(A|B) = 2/4 = .50
Let's define event C thus:
C: Observe T on first toss
Show that P(A|C) = 1/4 = .25
P(A|B) = P(A and B) ÷ P(B)