## Introduction to Probability

Aims: By the end of this chapter you will be able to calculate simple probability and solve problems using Tree Diagrams and Lattice Diagrams.

1. Probability is the "likelihood" of some event that have random outcomes. Probability can be interpreted as the relative frequency of an event if we carry out the experiment infinitely many times.

2. Let us run an experiment in such a way that a series of independent (the outcome now is independent of previous outcomes) and identical trials are produced. One such experiment is the flipping of an unbiased coin (equal likelihood for a Head and a Tail). If Head is recorded, we said that event H has occurred. If Tail is observed, event T has occurred instead of H. [The choices of T and H are arbitrary, we could use any other symbols]

3. We will let N to represent the total number of trials and n(A) the number of times that the event A was observed. In this example, event A could be H or T.

4. The ratio n(A)/ N is called the relative frequency of the event A.

5. For large values of N we find that the ratio n(A)/N tends to a number called the probability of the event A, which we denote by P(A).

Theoretical probability is the type of probability that we can work out the value of a particular event by using mathematical reasoning in which the argument often is based on some form of symmetry. Many questions that you will encounter in this course are based on theoretical probability.

Axioms of Probabilty:

1. 0 ≤ P(A) ≤ 1
Thus, you should get very suspicious if you obtain a probability value that is either greater than 1 or a negative value.
2. P(∅) = 0 and P(U) =1.
If event A never occur then it has zero probability and U implies that the event A is a certainty. The empty set () and the universe (the set of all possibilities,U also denoted by x) are complementary. We write the complement of A as A'.
• Let A be event that it rains today. Then A' is the event that it does not rain today.
If P(A)= 0.30 then P(A')=1-0.30 =0.70.
That is if the probability of raining today is 0.30 (30%) then the probability that it does not rain today must be 0.70 (70%).
• Thus P(A') = 1 - P(A).
3. If A and B are both subsets of U and are mutually exclusive (that is, no member in A is also in B and vice versa), then

P( A ∪ B ) = P(A) + P(B)

Example 1.
What is the probability of obtaining a 6 from a fair die?
Solution: P(6) = 1/6. A die can only have 6 possible outcomes and "6" is one of the six.

Example 2.
A card is selected from a pack of 52 cards. Find the probability that a selected card is

1. red
2. a heart
3. a 7
4. a red 7.
Solutions:
1. Half of this pack is red. So p(red) = 1/2.
2. A pack of card has 13 hearts ♥, 13 clubs ♣, 13 diamonds ♦, and 13 spades ♠. Thus, P(heart) = 13/52 = 1/4.
3. There are 4 sevens in the pack so P("7")=4/52 = 1/13.
4. There are 4 sevens and only two are reds. P(red '7')= 2/52 = 1/26.

Example 3.
Two fair coins are tossed. Find the probabilities that we obtain

2. one tail
3. at least one tail
4. a tail and a head
5. tail on the first coin and head on the second coin.
Solutions:
Method 1: by listing
The universe (all possible outcomes)= {HH, HT, TH, TT} where H represents head and T represents tail. HT means the first coin is head and the second is tail. A fair coin means that we have equal probability obtaining Head or Tail from the coin, i.e. P(H)=P(T) =1/2.
1. P(HH) = 1/4 from the list.
2. To obtain one tail we could have HT or TH. We use addition to take into account these two possible outcomes. Thus, P(one tail) = P(HT) + P(TH) . Note: "or" is addition (+) of one probability to another.
P(one tail)= 1/4 + 1/4 =1/2.
3. Outcomes HT or TH or TT satisfy the condition that at least one tail is showing. So P(at least one tail) = 3/4.
4. Here we have two possibilities HT or TH. So P(a tail and a head)=2/4 =1/2.
5. Note that this is not the same question as in (d). Here the order is important. The only possible outcome is TH. HT does not satisfy the requested condition.
P(TH) = 1/4.
Method 2: Tree diagram We could construct a tree diagram. The point on the left represents the state before the start of the experiment. It appears that for the sake of analysis tossing two coins simulataneuosly is equivalent to tossing one coin twice. Thus, the first toss is shown by the first two branches in the left. The end of these branches are the outcomes which is either H or T. The numbers represent probabilities. At the end of each branch is the start of the next toss. Again we have two possibilities, H or T and these outcomes are independent of the outcome of first toss. Thus, the P(H)=P(T)=1/2 on the second toss. The probability P(HT) for example is obtained by multiplying the numbers on the relevant branches. Thus, P(HT) = 1/2 x 1/2 = 1/4.
1. P(HH)=1/4 from the diagram.
2. P(one tail) = P(HT) + P(TH) = 1/4 + 1/4 from the diagram.
P(one tail) = 1/2 . From the tree diagram, identify all the possibilities with a tail in it and add up all these possibilities.
3. P(at least one tail) . We need to identify all possibilities with at least one tail in it and add up all these possibilities. p(at least one tail) =3/4.
4. From the tree, we have P(HT) and P(TH). So P(a tail and a head)=1/2.
5. P(TH)=1/4 from tree.
This tree diagram will prove more useful when you have more than two experiments and each experiment has more than 2 outcomes.

Example 4.
A box has 3 oranges, 2 apples and 5 pears. Find the probability that a selected fruit from the box is

1. orange
2. pear
3. NOT an orange.
Solutions:
1. P(orange)=3/10 .
2. P(pear) = 5/10 = 1/2.
3. P(not an orange) = 1 - P(orange)
P(not an orange) = 7/10. Note: This is a question on complementary.

Example 5.
We will roll two fair dice.

1. What is the probability that the sum of these dice is 5?
2. What is the probability that the sum of these dice is odd?
3. What is the probability that the sum of these dice is a factor of 10?
Solutions: A fair die means that we have equal probability obtaining 1, 2, 3, 4, 5 and 6. We could start by having a modification of a lattice diagram. (A lattice diagram will be similar to the diagram shown here with dots replacing all the numbers in the 6 by 6 array.) It lists all the possible sums of these two dice.
1. P(sum=5) = 4/36 = 1/9
as shown in the diagram. There are 36 possiblities in all and 4 of them have sum=5.
2. P(sum=odd) = 18/36 = 1/2
as shown in the diagram. We have to count all odd sums.
3. Factors of 10 are 1, 2, 5, and 10. So we need to count how many sums are 1, 2, 5, and 10. Note sum=1 is impossible so we are left with sum=2, sum=5 and sum=10.
P(sum = factor of 10) = 8/36 = 2/9.

Exercise