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

The probability of event " not A " is the difference between one and the probability of A.

Symbolically: **P(A ^{/} ) = 1 - P(A)**; where 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,}

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.

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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| rules of probability | Venn diagrams | probabiltiy and odds | practice | solutions |

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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

**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

and the **odds against A** is the **ratio of** **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%.

| rules of probability | Venn diagrams | probabiltiy and odds | practice | solutions |

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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.

___________________________

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

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 |

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| rules of probability | Venn diagrams | probabiltiy and odds | practice | solutions |

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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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