Skip to main content

6. Probability of an Event

Definition of a Probability

Suppose an event E can happen in r ways out of a total of n possible equally likely ways.

Then the probability of occurrence of the event (called its success) is denoted by

`P(E)=r/n`

The probability of non-occurrence of the event (called its failure) is denoted by

`P(barE)=(n-r)/n=1-r/n`

Notice the bar above the E, indicating the event does not occur.

Thus,

`P(barE)+P(E)=1`

In words, this means that the sum of the probabilities in any experiment is `1`.

Definition of Probability using Sample Spaces

When an experiment is performed, we set up a sample space of all possible outcomes.

In a sample of N equally likely outcomes we assign a chance (or weight) of `1/N` to each outcome.

We define the probability of an event for such a sample as follows:

The probability of an event E is defined as the number of outcomes favourable to E divided by the total number of equally likely outcomes in the sample space S of the experiment.

That is:

`P(E)=(n(E))/(n(S)`

where

  • `n(E)` is the number of outcomes favourable to E and

  • `n(S)` is the total number of equally likely outcomes in the sample space S of the experiment.

Properties of Probability

(a) 0 ≤ P(event) ≤ 1

In words, this means that the probability of an event must be a number between `0` and `1` (inclusive).

(b) P(impossible event) = 0

In words: The probability of an impossible event is `0`.

(c) P(certain event) = 1

In words: The probability of an absolutely certain event is `1`.

Example 1

What is the probability of...

(a) Getting an ace if I choose a card at random from a standard pack of `52` playing cards.

Answer

In a standard pack of 52 playing cards, we have:

2 3 4 5 6 7 8 9 10 J Q K A
2 3 4 5 6 7 8 9 10 J Q K A
2 3 4 5 6 7 8 9 10 J Q K A
2 3 4 5 6 7 8 9 10 J Q K A

There are 4 aces in a normal pack. So the probability of getting an ace is:

`P("ace")=4/52 = 1/13`

Get the Daily Math Tweet!
IntMath on Twitter

(b) Getting a `5` if I roll a die.

Answer

6-sided gaming die (dice)
Image source

A die has 6 numbers.

There is only one 5 on a die, so the probability of getting a 5 is given by:

`P(5)=1/6`


(c) Getting an even number if I roll a die.

Answer

Even numbers are `2, 4, 6`. So

`P("even")=3/6=1/2`

Easy to understand math videos:
MathTutorDVD.com

(d) Having one Tuesday in this week?

Answer

Each week has a Tuesday, so probability = `1`.

Easy to understand math videos:
MathTutorDVD.com

Example 2

There are `15` balls numbered `1` to `15`, in a bag. If a person selects one at random, what is the probability that the number printed on the ball will be a prime number greater than `5`?

Answer

The primes between `5` and `15` are: `7, 11, 13`.

So the probability `=3/15=1/5`

Please support IntMath!

Example 3

The names of four directors of a company will be placed in a hat and a 2-member delegation will be selected at random to represent the company at an international meeting. Let A, B, C and D denote the directors of the company. What is the probability that

(a) A is selected? (b) A or B is selected? (c) A is not selected?

Answer

The possible outcomes are: AB, AC, AD, BC, BD, CD.

[There are a few explanations for each answer - hopefully at least one of them makes sense!]

Part (a)

Explanation 1: The probability is `3/6=1/2` since when we choose A, we must choose one of the remaining 3 directors to go with A. There are `C_2^4=6` possible combinations.

Explanation 2: Probability that A is selected is `{C_1^1 times C_1^3}/{C_2^4} = 3/6 = 1/2`

[Choose A (`C_1^1`), and then choose one from the 3 remaining directors (`C_1^3`), divided by the number of possible outcomes: `C_2^4`.]

Part (b)

Explanation 1: The probability of getting A or B first is `2/4=1/2`.

Now to consider the probability of selecting A or B as the second director. In this case, the first director has to be C or D with probability `2/4` (2 particular directors out of 4 possible).

Then the probability of the second being A or B is `2/3` (2 particular directors out of the remaining 3 directors).

We need to multiply the two probabilities.

So the probability of getting A or B for the second director is `2/4 xx 2/3 = 1/3`

The total is: `1/2 + 1/3 = 5/6`

Explanation 2: Probability that A or B is selected is

`frac{C_1^1 times C_1^3 + C_1^1 times C_1^2}{C_2^4}` `=frac{3+2}{6}` `=5/6`

[Choose A as above, then choose B from the remaining 2 directors in a similar way.]

Explanation 3: If A or B is chosen, then we cannot have the case C and D is chosen. So the probability of A or B is given by:

`P("A or B") = 1-P("C and D")` `=1-1/6` `=5/6`

Part (c)

Probability that A is not selected is `1-1/2=1/2`

Extension

Consider the case if we are choosing 2 directors from 5. The probabilities would now be:

(a) Probability that A is selected is

`frac{C_1^1 times C_1^4}{C_2^5}=4/10=2/5`

[Choose A (`C_1^1`), and then choose one from the 3 remaining directors (`C_1^4`), divided by the number of possible outcomes: `C_2^5`.]

(b) Probability that A or B is selected is

`frac{C_1^1 times C_1^4 + C_1^1 times C_1^3}{C_2^5}` `=frac{4+3}{10}` `=7/10`

[Choose A as above and then choose B from the remaining 3].

(c) Probability that A is not selected is `1-2/5=3/5`.

Please support IntMath!

Coming next...

2 3 4 5 6 7 8 9 10 J Q K A
2 3 4 5 6 7 8 9 10 J Q K A
2 3 4 5 6 7 8 9 10 J Q K A
2 3 4 5 6 7 8 9 10 J Q K A

The next 2 sections give more examples of probability:

Singapore Toto

Poker

top

Search IntMath, blog and Forum

Search IntMath

Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Math Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!


See the Interactive Mathematics spam guarantee.