Data

Basic Laws of Probability

The Compliment Rule

When dealing with probabilities, the sample space is 100% or 1.
The entire sample space is made up of A +

.
Anything in the sample space is either in A or

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For Any event A,

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The last formula also tells us that

In sets,       is all elements in A plus all elements in B.
   is everything inside the rings in the Venn diagram below.

   is all elements that are not in A and not in B.

   is everything outside the rings in the Venn diagram below.

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Example 1:

If the probability of event A occurring is .6,

Find

Solution:
Since P(A) = .6

1 - .6

Therefore the solution is .4. ___________________________________________________

The Union Rule

For any events A & B:

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Example 2:

If P(A) = .53,

P(B)= .72

Find the following:

A) The probability that A does not occur.

Solution:

1 - .53

.47

Therefore the solution is .47.

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B) The probability that A or B occurs.

Solution:

.53 + .72 - .48

.77

Therefore the solution is .77.

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C) The probability neither A nor B occurs.

Solution:
This is not A and not B, which is the outside of the rings, or

Anything complemented is always one minus the uncomplemented object.

1 - .77

.23

Therefore the solution is .23.

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Mutually Exclusive Events

If A and B are events that have no common outcomes, then A & B are Mutually Exclusive and

If A and B are mutually exclusive, then they can be represented by the following Venn Diagram.

If A & B are mutually exclusive events then:

Since the intersection of A and B is empty, there is nothing to subtract.

Student #1:
Determine if the two events A & B are mutually exclusive.

Event A: The sum of rolling two dice four times is 1, 3, 5, 9.
Event B: The sum of rolling two dice four times is 4, 5, 10, 11.

Answer

Student #2:
Determine in the two events A & B are mutually exclusive.

Event A: A card drawn from a deck of cards is Black.
Event B: A card drawn from a deck of cards is a diamond.

Answer

Student #3:
Determine in the two events A & B are mutually exclusive.

Roll two dice:
Event A: A die comes up a 3.
Event B: The sum of the dice is 10.

Answer

The Union Rule for Mutually Exclusive Events

If all pairs of the events are mutually exclusive,

then

When any event intersected with any other event is the empty set, the union is the sum of the probabilities for each event.

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Example 3:

At a local bank the probability of having various number of teller lanes open are given:

#of Lanes Open Probability
1 .18
2 .21
3 .12
4 .26
5 .23

What is the Probability that at least 2 teller lanes are open at any given random time?

#of Lanes Open	Probability
1	.18
2	.21
3	.12
4	.26
5	.23

Solution:
Let        be the event that exactly one teller lane is open.
Let        be the event that exactly two teller lanes are open.
Let        be the event that exactly three teller lanes are open.
Let        be the event that exactly four teller lanes are open.
Let        be the event that exactly five teller lanes are open

The question is:    are these events all mutually exclusive with each other?

Ask yourself this question.    Can you have exactly one teller lane open, and have exactly two teller lanes open at the same time?    The answer is no, so they are mutually exclusive events.
Since all pairs of events are mutually exclusive, i.e.

, etc.,

We can use the union rule for mutually exclusive events.

P(at least 2 lanes are open)

= .21 + .12 + .26 + .23

= .82
Therefore the solution is .82.

Or use the complement:

P(at least 2 lanes are open) = 1 - P(1 lane open)

= 1 - .18

= .82
Therefore the solution is .82.
_

Example 4:
Consider the sample space for flipping a coin twice:

S = {HH,HT,TH,TT} where HH is Heads Heads, HT is Heads Tails, etc.

Let E be the event "the same side of the coin shows on both flips".
{HH,TT}

Let F be the event "the second flip is tails".
{HT, TT}

Describe the following in words, and find each sample space.

A)
Solution:

The same side of the coin does not show on both flips.
(One head and one tail.)

B)
Solution:

The second flip is not tails.
(The second flip is heads.)

C)
Solution:

The same side of the coin shows on both flips, and the second flip is tails.
(Both flips are tails.)

D) Are E and F mutually exclusive?

Solution:
No, since        is not empty, E and F are NOT mutually exclusive.

E)
Solution:

The same side of the coin shows on both flips, or the second flip is tails.

Student #4:
Five women at a hotel give their jackets to the valet for cleaning.    If the jackets are randomly handed back to the owners, what is the probability that at least one women receives the wrong jacket?

.56 .87 .99 .45

Answer

Student #5:
Data from Parham county shows there is a .62 probability of listening to Kcob 96 fm, and a .54 probability of listening to Coast 109.    If the probability of listening to both stations is .24, what is the probability that a person selected at random does not listen to either station?

.65 .08 .25 .55

Answer

Example 5:

A dad has 8 red candies, and 9 blue candies.    If 6 candies are to be chosen at random, what is the probability that:

A)    At least one will be blue.
Solution:
The sample space for blue candies is:

Remember, 6 candies total must be chosen.

S={0B6R, 1B5R, 2B4R, 3B3R, 4B2R, 5B1R, 6B0R}
where (1B5R) means 1 blue and 5 red

P(at least one blue) =

P(1B5R) + P(2B4R) + P(3B3R) + P(4B2R) + P(5B1R) + P(6B0R)

=.998
Therefore the solution is .998.

Or use the complement.
The complement is    1 - (the unused portion of the sample space).
P(at least one blue) = 1 - P(0B6R) =

Therefore the solution is .998.

B)    No more than two will be red.
Solution:
The sample space for red candies is:

S={0R6B, 1R5B, 2R4B, 3R3B, 4R2B, 5R1B, 6R0B}

P(no more than two)

= P(0R6B) + P(1R5B) + P(2R4B)

Therefore the solution is .37.

Or use the complement.

P(no more than two)

= 1 - [P(3R3B) + P(4R2B) + P(5R1B) + P(6R0B)]

= .37

Therefore the solution is .37.

Homework Problems

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Use the subject D2.4

Student #1:    The answer is B:

Solution:
Both events have a common outcome, namely the 5. Since    the two events are NOT mutually exclusive.

Go Back to Lesson

Student #2:    The answer is A:

Solution:
If you draw one card from a deck of cards, can it be a black card and a diamond?    No, these two events can not happen at the same time, so they are mutually exclusive events.

Therefore the events are mutually exclusive.

Go Back to Lesson

Student #3:    The answer is A:

If A & B are not ME, then they will be able to happen at the same time.    This means a 3 is rolled, AND the sum is 10.

(3, ?),    3 + ?    has a sum of 10?

This requires the number 7 on the second dice, but 7 is not on a dice.

Therefore A & B are mutually exclusive events.

Go Back to Lesson

Student #4:    The answer is C:

Solution:

The sample space for getting the wrong jacket is:

S={0 wrong, 1 wrong, 2 wrong, 3 wrong, 4 wrong, 5 wrong}

We want at least one wrong jacket which is:

P(1 wrong) or P(2 wrong) or P(3 wrong) or P(4 wrong) or P(5 wrong)

Remember the word or means add.
P(1) + P(2) + P(3) + P(4) + P(5)

This requires quite a bit of work.    What is the only thing in the sample space that is not used?    The unused part of the sample space is P(0), which is zero wrong jackets.

So you can either calculate:

P(1) + P(2) + P(3) + P(4) + P(5), or use the complement    1 - P(0).

Find the sample space?    |S| = ?

There are 5! ways to give out 5 jackets.    Since the first person has 5 ways to receive a jacket, the second person then has 4 ways to receive a jacket, etc.

There is only 1 way to return no wrong jackets.    If there are no wrong jackets, that means everyone has the right jacket.    The 1 way to do that is to give everyone the correct jacket.    So, there are 1/5! ways for no wrong jackets to occur.

P(at least one wrong jacket) = 1 - P(no wrong jackets)

=

= .99

Therefore the solution is .99.

Go Back to Lesson

Student #5:    The answer is B:

Solution:
Let A be the event of listening to Kcob 96

Let B be the event of listening to Coast 109

The probability of not listening to A or B is:

= 1 - (.62 + .54 - .24)

= .08

Therefore the solution is .08.
Go Back to Lesson

Data Basic Laws of Probability

The Compliment Rule

The Union Rule

Answer

Answer

Answer

The Union Rule for Mutually Exclusive Events

Answer

Answer

Data

Basic Laws of Probability