6. Probability Of An Event - Interactive Mathematics

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3. 6. Probability of an Event

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• Counting and Probability - Introduction
• 1. Factorial Notation
• 2. Basic Principles of Counting
• 3. Permutations
• 4. Combinations
• 5. Introduction to Probability Theory
• 6. Probability of an Event
• Singapore TOTO
• Probability and Poker
• 7. Conditional Probability
• 8. Independent and Dependent Events
• 9. Mutually Exclusive Events
• 10. Bayes’ Theorem
• 11. Probability Distributions - Concepts
• 12. Binomial Probability Distributions
• 13. Poisson Probability Distribution
• 14. Normal Probability Distribution
• The z-Table
• Normal Distribution Graph Interactive

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6. Probability of an Event

Definition of a Probability

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Definition using Sample Spaces Properties of 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`

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

Answer

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`

(d) Having one Tuesday in this week?

Answer

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

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`

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`.

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

5. Introduction to Probability Theory Singapore TOTO

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