Probability – SS2 Mathematics Lesson Note (Third Term)

Probability is a measure of the likelihood that an event will occur in any one trial. It can also be defined as the ratio of the required outcome to the total outcome. Probability can be applied in several areas, such as insurance and industrial quality control.

Behavioural objectives

By the end of the lesson, students should be able to:

• define probability and state its range of values;
• find the probability of an event from the sample space;
• list the sample spaces for coins and dice and use them to find probabilities;
• use the addition rule for mutually exclusive events;
• use the multiplication rule for independent and dependent events, with and without replacement.

How is probability calculated?

Probability = number of required outcomes / total number of outcomes

Key terms

1. Sample space: the set of all possible outcomes of a random experiment. It is denoted by S, and the number of outcomes is denoted by n(S).
2. Event space: the collection of outcomes of a random experiment that we want. The number of these outcomes is denoted by n(E).
3. Outcome: the result of an experiment in probability.

What is the range of probability?

The probability that an event is certain to happen is 1, while the probability that an event is certain not to happen is 0. The range of probability is therefore 0 ≤ Pr(E) ≤ 1.

Hence, Pr(an event will occur) + Pr(an event will not occur) = 1.

Example: balls in a box

From a box containing 2 red, 6 white and 5 black balls, a ball is randomly selected. What is the probability that the selected ball is: (i) black, (ii) white, (iii) not black?

Solution:

n(S) = 2 + 6 + 5 = 13

n(red) = 2, n(white) = 6, n(black) = 5

1. Pr(black) = 5/13
2. Pr(white) = 6/13
3. Pr(not black) = 1 – Pr(black) = 1 – 5/13 = 8/13

What is an equiprobable sample space?

Coins

A coin has two faces called the head (H) and the tail (T). The outcome of an experiment involving a coin depends on the number of trials.

• In a single throw of a fair coin: {H, T}, so n(S) = 2.
• Two coins thrown once, or a coin thrown twice: {HH, HT, TH, TT}, so n(S) = 4.
• Three coins thrown once, or a coin thrown three times: {HHH, HHT, HTT, HTH, THH, THT, TTH, TTT}, so n(S) = 8.

Example: In a single throw of two fair coins, find the probability that: (i) two tails appear, (ii) one head and one tail appear, (iii) two heads appear, (iv) one tail and one head appear in that order.

Solution:

S = {HH, HT, TH, TT}, n(S) = 4

1. Pr(two tails) = {TT} = 1/4
2. Pr(one head, one tail) = {HT, TH} = 2/4 = 1/2
3. Pr(two heads) = {HH} = 1/4
4. Pr(one head, one tail in that order) = {HT} = 1/4

Dice

A fair die is a six-faced die. A die could be tossed in a different number of trials.

When a die is tossed once: {1, 2, 3, 4, 5, 6}, so n(S) = 6.

When a die is tossed twice, or two dice are tossed, the outcomes are shown in the table below. The total number of outcomes is 36.

Example: If two fair dice are tossed together, what is the probability that the total score will be: (i) ten, (ii) at least 4, (iii) a prime number?

Solution: Use the table above.

1. Total score of 10 = (4,6), (5,5), (6,4) = 3 outcomes.
   Pr(total score of 10) = 3/36 = 1/12
2. Total score of at least 4 = (1,3), (1,4), (1,5), (1,6), (2,2), (2,3) … = 33 outcomes.
   Pr(total score of at least 4) = 33/36 = 11/12
3. Total score of a prime number = (1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5) = 15 outcomes.
   Pr(total score of a prime number) = 15/36 = 5/12

What are mutually exclusive events?

Two events are mutually exclusive if they cannot occur at the same time. That is, there is no common element between them. This leads to the addition rule.

Addition rule: If E₁, E₂, E₃, … are mutually exclusive events, then Pr(E₁ ∪ E₂ ∪ … ∪ E_n) = Pr(E₁) + Pr(E₂) + … + Pr(E_n).

Words such as “or” and “either” are used to indicate the addition of probabilities.

Example: In a single throw of a fair die, what is the probability that an even number or a perfect square greater than 1 shows up?

Solution:

S = {1, 2, 3, 4, 5, 6}, n(S) = 6

Even number = {2, 4, 6}, so Pr(even number) = 3/6

Perfect square greater than 1 = {4}, so Pr(perfect square greater than 1) = 1/6

The number 4 is in both sets, so these two events are not mutually exclusive and the numbers cannot simply be added. The required outcomes are {2, 4, 6}.

Pr(even number or perfect square greater than 1) = 3/6 = 1/2

What are independent and dependent events?

Independent events

Two or more events are independent when the occurrence of one event does not affect the occurrence of the other events in any way. Hence, the events can occur independently. For example, obtaining a 6 in a single throw of a die and obtaining a tail in a toss of a coin. This leads to the multiplication rule.

Multiplication rule: If E₁, E₂, E₃, … E_n are independent events, then Pr(E₁ ∩ E₂ ∩ … ∩ E_n) = Pr(E₁) × Pr(E₂) × … × Pr(E_n).

When two events are combined with words such as “and” and “both”, the probabilities of the events are multiplied.

Events are independent when picking is done with replacement.

Dependent events

Two or more events are dependent when the occurrence of one event affects the occurrence of the other event(s). Events are dependent when picking is done without replacement.

Conditional probability

Conditional probability is an application of the multiplication rule.

Example: students selected for a debate

A class consists of 8 men and 7 ladies. Two students are selected randomly to represent the class in a debate. Find the probability that the two students selected are: (a) both ladies, (b) both men, (c) of the same sex, (d) of different sexes.

Solution:

Total students = 8 + 7 = 15

n(men) = 8, n(ladies) = 7

• Pr(both ladies) = 7/15 × 6/14 = 1/5
• Pr(both men) = 8/15 × 7/14 = 4/15
• Pr(same sex) = Pr(M and M) or Pr(L and L) = 4/15 + 1/5 = 7/15
• Pr(different sexes) = Pr(M and L) or Pr(L and M) = (8/15 × 7/14) + (7/15 × 8/14) = 4/15 + 4/15 = 8/15

Example: marbles in a box

Two marbles are selected randomly, one after the other without replacement, from a box containing 5 red, 7 white and 8 blue marbles. Find the probability that: (a) both are red, (b) one is white and the other is blue, (c) both are of the same colour.

Solution:

Total marbles = 5 + 7 + 8 = 20

n(R) = 5, n(W) = 7, n(B) = 8

• Pr(both red) = Pr(RR) = 5/20 × 4/19 = 1/19
• Pr(one white, one blue): the arrangement is important.
  Pr(WB) or Pr(BW) = (7/20 × 8/19) + (8/20 × 7/19) = 14/95 + 14/95 = 28/95
• Pr(same colour) = Pr(RR) or Pr(WW) or Pr(BB)
  = (5/20 × 4/19) + (7/20 × 6/19) + (8/20 × 7/19)
  = 20/380 + 42/380 + 56/380
  = 118/380 = 59/190

Evaluation

1. A box contains 2 red, 6 white and 5 black balls. A ball is picked at random. Find the probability that it is (a) red, (b) not red.
2. In a single throw of two fair coins, find the probability that two heads appear.
3. How many outcomes are there when two fair dice are tossed together?
4. State the addition rule and the words that show that probabilities should be added.
5. Explain the difference between independent and dependent events.

Assignment

1. Two fair dice are tossed. Find the probability (a) of not getting a total of 9, (b) that the two dice show the same number.
2. In a single throw of three fair coins, find the probability that: (a) one head and two tails appear, (b) at least one head appears.
3. A box contains 2 white and 3 blue identical marbles. If two marbles are picked at random, one after the other, without replacement, what is the probability of picking two marbles of different colours?
4. A ball is picked at random from a bag containing 5 green balls, 3 white balls and 2 black balls. What is the probability that it is either green or black?