probability rules, odds

Probability: Rules, Odds

1) The probability of event A is a positive number greater than or equal to 0
but less than or equal to 1. Symbolically: 0 [ P(A) [ 1

2) The sum of all probabilities = 1. Symbolically: S P(A) = 1

3) Complement Rule: ( not A or A ^/ is complement of A)

The probability of event " not A " is the difference between one and the probability of A.
Symbolically: P(A ^/ ) = 1 - P(A); where A ^/ is "not A"
(some books use ~A -- the tilde (~) -- for "not A")

Union and Intersection

The UNION of sets A and B contains all the elements in either set A or set B.

We use the U symbol ( 4 ) for UNION. Symbolically: (A or B) = (A 4 B).

Example: If A = {1, 2, 3, 4} and B = {3, 4, 5, 6} then (A 4 B) = {1, 2, 3, 4, 5, 6}

The INTERSECTION of sets A and B contains all the elements in both set A and set B.

We use the inverted U symbol ( 3 ) for INTERSECTION. Symbolically: (A and B) = (A 3 B)

Example: If A = {1, 2, 3, 4} and B = {3, 4, 5, 6} then (A 3 B) = {3, 4,}

4) General Addition Rule:

If we know the probabilities of events A and B, to find the probability that either event occurs, we use the general addition rule. It says that the probability of either A OR B is the sum of the probabilities less the probability of both events happening. (the intersection)

Symbolically: P(A or B) = P(A) + P(B) - P(A 3 B)

If events A and B are mutually exclusive, they can't both happen, then
P(A 3 B) = 0 so P(A 4 B) = P(A) + P(B)

For example, a baseball player can't hit a home run and strike out in the same turn at bats. These events are mutually exclusive -- the occurrence of one excludes the other.

Venn Diagrams

We often use Venn diagrams to depict the unions and intersections of set.
Here, the blue circle represents the probability that event A occurs.
The red circle represents the probability that event B occurs.

Figure 1, P(A 3 B) = 0. Events A and B are mutually exclusive: P(A 4 B) = P(A) + P(B)

Figure 2, (A 3 B) is the shaded area, containing elements in both sets.
When we add P(A) + P(B) we include this area twice.
That's why we subtract the intersection from the sum: P(A 4 B) = P(A) + P(B) - P(A 3 B).

Examples:

1) If A and B are mutually exclusive events and P(A) = 0.30 and P(B) = 0.55
a) the probability that either A or B occurs is 0.30 + 0.55 = 0.85
b) the probability that neither A nor B occurs is 1 - (0.30 + 0.55) = 0.15

2) If P(A) = 0.30 and P(B) = 0.55 with P(A 3 B) = 0.15

a) the probability that either A or B occurs is 0.30 + 0.55 - 0.15 = 0.70
b) the probability that neither A nor B occurs is 1 - (0.70) = 0.30
c) the probability that A occurs but B doesn't is 0.30 - 0.15 = 0.15

d) the probability that B occurs but A doesn't is 0.55 - 0.15 = 0.40

Let's look at the Venn diagram of this one:

We start with the intersection: P(A 3 B) = 0.15 and since P(A) = 0.30,
we get the other 0.15 which is the probability that event A occurs but B doesn't.

Also, P(B) = 0.55 so 0.55 - 0.15 = 0.40 is the probability that B occurs but A doesn't.

The 0.30 outside the circles is the probability that neither A nor B occurs.

When we add up all the probabilities, we get one.

Example

We choose 200 students at random from Bored-to-Death High School.
There are 48-SecI, 42-SecII, 101-SecIII, 7-SecIV and 2-SecV students in our sample.
If we choose 1 student from the sample at random, what is the probability that s/he will be:

a) a SecI student? b) a SecIV student? c) either a SecI or a SecIV student?

solution:

a) Since there are 48 SecI students in 200, P(SecI) = 48 / 200 = 0.24

b) Since there are 7 SecIV students in 200, P(SecIV) = 7 / 200 = 0.035

c) The events are mutually exclusive so P(A 4 B) = P(A) + P(B)

P(SecI or SecIV) = (48 + 7)/ 200 = 55 / 200 = 0.275 which of course is 0.24 + 0.035

Probability and Odds

1) Changing Probability to Odds:

If p is the probability that event A occurs, then
the odds that A will occur are a / b, with

Probability is the ratio of successes to total number of trials -- x : n

Odds in favour of the event are the ratio of successes to failures if p > 0.50 -- x : (n - x)

Odds against the event are the ratio of failures to successes if p < 0.50 -- (n - x) : x.
where p is P(A), x is the number of successful outcomes from n trials of the experiment.

If the probability of rain today is 60%, the odds that it will rain are 3 : 2

WHY? since 60% = 3/5 which means 3 successes and 2 failures in 5 trials.

If the probability of rain today is 30%, the odds that it won't rain are 7 : 3

WHY? since 30% = 3/10 which means 3 successes and 7 failures in 10 trials. Because odds are stated with the larger number first, and 7 > 3, we say the odds against the event are 7 : 3.

Properties of Odds

Both a and b are always positive with a = p and b = 1 - p

Recall that 1 - p is the probability of not A.

So the odds in favour of event A is the ratio of P(A) : P(A^/ ) if P(A) > P(A^/ )

and the odds against A is the ratio of P(A^/ ) : P(A) if P(A^/ ) > P(A).

Odds are always reduced to lowest terms!! We don't say the odds are 9 to 6, we say 3 to 2.

Odds are stated with the larger number first. ie: a > b,

so if the odds of event A are 2 : 3, we say the odds against A are 3 : 2.

Examples

1) If the probability of event A is 0.36, then p = 0.36 and 1 - p = 0.64,
and since 1 - p > p, we say the odds against A are 64 : 36 which equals 16 : 9.

2) If P(W) -- the probability that a certain horse wins the race -- is 0.45, the odds
of winning are 45 : 55 or 9 : 11. Since 11 > 9, we say the odds against winning are 11: 9.

2) Changing Odds to Probability:

If the odds in favour of A are a : b, then
P(A) -- the probability of A -- is

Say the bookies are giving 3 to 2 odds in favour of the Leafs winning their next game. It means that if they played 5 games they won 3 and lost 2; if they played 10 games they won 6 and lost 4. Since probability is the number of successes divided by the total number of trials, we add the two numbers a and b to get the total number of trials -- the denominator. The probability that the Leafs will win their next game then is 3 / (3 + 2) = 3 / 5 or 60%.

Practice

1/ Convert each of the following probabilities to odds:

(a) The probability of rolling 10 or less with a pair of balanced dice is 11/12.

(b) If a researcher randomly selected 5 of 30 homes to be included in a study, the probability is 1/6 that any particular home will be included.

___________________________

2/ Convert each of the following odds to probabilities:

(a) The odds are 19 to 5 that a given horse will not win the Kentucky Derby.

(b) If 5 cards are drawn with replacement from an ordinary deck of 52 playing cards,
the odds are 13 to 3 that at most 3 of the cards will be red.

(c) If 2 persons are chosen at random from a group of 10 men and 12 women,
the odds are 40 to 37 that one man and one woman will be selected.

___________________________

3/ If a contractor feels that 17 to 8 are fair odds that a job will be finished on time,
what probability does he assign to the event?

___________________________

4/ In the diagram, G is the event that Harry graduates from college this year,
and C is the event that Harry's father buys him a car this year.
Explain in words what events are represented by the regions numbered:

a) 1 b) 2 c) 3 d) 4 e) 3 & 4 f) 2 & 3 g) 1 & 2 & 3

___________________________

5/ A company has discovered a way of evaluating the success of both their
radio and TV advertising. If R is the event that their radio advertising is successful
and T is the event that their television advertising is successful
and P(R) = 0.62, P(T) = 0.75, and P(R 3 T) = 0.43, find:

a) P(R^/ ) b) P(R 4 T) c) P(R 3 T ') d) P(R 4 T ') e) draw a Venn diagram

___________________________

Solutions

1/ a) odds are 11 to 1 of getting 10 or less.

b) odds are 5 to 1 that the home is not included.

c) odds are 13 to 3 of getting at least 2 heads.

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2/ a) Since P(win) = 5/24, then P(not win) = 1 - P(W) = 19/24 = 0.7917

b) P(Red) = 13/16 = 0.813

c) P(1 man, 1 woman) = 40/77 = 0.52

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3/ P(F) = 17/25 = 0.68

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4/ a) Harry graduates but doesn't get a car.

b) Harry graduates and gets a car.

c) Harry gets a car but doesn't graduate.

d) Harry doesn't graduate and he doesn't get a car.

e) Harry doesn't graduate this year.

f) Harry gets a car this year.

g) Harry graduates and/or gets a car.

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5/ P(R) = 0.62, P(T) = 0.75, and P(R 3 T) = 0.43, find:

a) P(R') = [ 1 - P(R) = 0.38 ] b) P(R 4 T) = [P(R) + P(T) - P(R 3 T) = 0.94 ]

c) P(R 3 T^/ ) = [P(R) - P(R 3 T) = 0.19 ] d) P(R 4 T^/ )= [P(R) + P(T^/ ) - P(R 3 T^/ )= 0.68 ]

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Back to Statistics MathRoom Index

a) P(R') = [ 1 - P(R) = 0.38 ]	b) P(R 4 T) = [P(R) + P(T) - P(R 3 T) = 0.94 ]
c) P(R 3 T^/ ) = [P(R) - P(R 3 T) = 0.19 ]	d) P(R 4 T^/ )= [P(R) + P(T^/ ) - P(R 3 T^/ )= 0.68 ]