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")

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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,}

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.

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.

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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

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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.

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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%.

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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.

(c) The probability of getting at least 2 heads in five flips of a balanced coin is 13/16.

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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.

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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?

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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

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5/ A company has discovered a way of evaluating the success of both their
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

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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 ]

e)

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(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - 2006).