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.

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

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

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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.

Even numbers are 2, 4, 6. So

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

(d) Having one Tuesday in this week?

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?

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?

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.

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