Counting & Probability
1. Basic Probability
1. Experiment: is a situation involving chance or probability that leads to results.
2. Outcome: is the result of a single trial of an experiment.
3. Event: is one or more outcomes of an experiment.
4. Probability of an event happening is: (number of favorable cases)/(number of possible cases)
A spinner has 4 equal sectors colored yellow, blue, green and red. After spinning the spinner, what is the probability of landing on each color?
Outcomes: The possible outcomes of this experiment are yellow, blue, green, and red.
Probabilities: P(yellow) = 1/4, P(blue) = 1/4, P(green) = 1/4, P(red) = 1/4.
Experiment 2: (Dice)
A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?
Outcomes: The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.
Probabilities: P(1) = 1/6, P(2) = 1/6, P(3) = 1/6, P(4) = 1/6, P(5) = 1/6, P(6) = 1/6, P(even) = 3/6 = 1/2, P(odd) = 3/6 = 1/2
Experiment 3: (Jar of Marbles)
A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?
Outcomes: The possible outcomes of this experiment are red, green, blue and yellow.
Probabilities: P(red) = 6/22 = 3/11, P(green) = 5/22, P(blue) = 8/22 = 4/11, P(yellow) = 3/22.
In Experiments 1 & 2 the probability of each outcome is always the same. However, in experiment 3, the outcomes are not equally to occur. You are more likely to choose a blue marble than any other color. You are least likely to choose a yellow marble.
E = An Event
n(E) = Number of outcomes of event E
Let E1 and E2 be mutually exclusive events (meaning that there are no common outcomes).
Let event E describe the situation where either event E1 or event E2 will occur.
The number of times event E will occur can be given by the expression:n(E) = n(E1) + n(E2)
n(E) = Number of outcomes of event E
n(E1) = Number of outcomes of event E1
n(E2) = Number of outcomes of event E2
Suppose that event E1 can result in any one of n(E1) possible outcomes; and for each outcome of the eventE1, there are n(E2) possible outcomes of event E2.
Together there will be n(E1) × n(E2) possible outcomes of the two events.
That is, if event E is the event that both E1 and E2 must occur, then
n(E) = n(E1) × n(E2)
What is the total number of possible outcomes when a pair of coins is tossed?
The events are described as:
E1 = toss first coin (2 outcomes, so n(E1) = 2.)
E2 = toss second coin (2 outcomes, so n(E2) = 2.)
They are independent, since neither toss affects the outcome of the other toss.
So n(E) = n(E1) × n(E2) = 2 × 2 = 4
[We could list the outcomes: HH HT TH TT].
3. Factorial Notation
n factorial is defined as the product of all the integers from 1 to n (the order of multiplying does not matter) .
We write "n factorial" with an exclamation mark as follows: n!n! = (n)(n − 1)(n − 2)...(3)(2)(1)
The value of 0! is 1. (It is a convention)
Let us take an example:
4! = 4 × 3 × 2 × 1 = 24
The number of ways (or combinations) in which r objects can be selected from a set of n objects, where repetition is not allowed, is denoted by nCr = (n!) / [r! (n-r)!], or:
Consider 4 students walking toward their school entrance. How many different ways could they arrange themselves in this side-by-side pattern?
The number of different arrangements is 4! = (4)(3)(2)(1) = 24. There are 24 different arrangement, or permutations, of the four students walking side-by-side.
The notation for a permutation is: nPr. (The symbol nPr is the number of permutations formed from n objects taken r at a time)
The formula for a permutation is:
5P3 = (5!) / (5-3)! = 60
The manager can arrange them in 60 ways
However, in the above example, repetitions are not allowed!
What if we had repetitions?
In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical.
The equation for permutations with repetitions is:
In how many ways can the six letters of the word "mammal" be arranged in a row?
The word mammal consists of 3 "m"s and 2 "a"s, therefore we use the above formula,
(6!) / [(3!)(2!)] = 60
The six letters of the word mammal can be arranged in 60 ways.