Recall that an event is any subset of the sample space, If A and B are simple events then a compound event is any event combining two or more simple events

Casper College contracts with Wyoming Pen and Ink Corp. to produce ball point pens that will be used as promotional material. The two most common defects of ballpoint pens are that ink does not flow and the clicker does not work. Concerned that they might offend prospective students or contacts in the community, the college is concerned about the probability that if a pen is selected at random it is defective.

The event of the pen being defective is a compound event. If your experiment is selecting one pen at random from a batch of pens and you have the following events

A: ink does not flow in the pen

B: clicker does not work in the pen

Then P (defective pen is selected) = P(A or B)

Note about or - this is an inclusive or that means

P (A or B)= P(A occurs or B occurs or they BOTH occur)

Now let's take this example a little further - suppose the company that produces the ball point pens tests 1000 pens and finds that in 27 pens ink did not flow and in 13 pens the clicker did not work and in 9 pens both of these events occurred. What is the probability of?

Ink not flowing: P (A) = (27)/1000 = .027

Clicker not working: P (B) = (13)/1000 = .013

Both ink not flowing and clicker not working: P (A and B) = 9/1000 = .009

What is the probability of a defective pen - we use the rule called the addition rule:

So for our example:

P (A or B) = P (A) + P (B) - P (A and B) = .027 + .013 - .009 = .031

If the company produces 1500 pens for CC - how many do you expect to be defective? (.031)*(1500) = 46.5, so say 47

We can state the addition rule intuitively as follows:

EXPERIMENT: Draw a card from a deck of card. Here are some events

A: getting a heart

B: getting a king

C: getting a black card

D: getting a 7

Note that P (A) = 1/4, P (B) = 1/13, P(C) = 1/2, P (D) = 1/13

Find P (A or B)

We can use the formula: P (A or B) = P (A) + P(B) - P (A and B) and the fact that P (A and B) means selecting a heart and a king - since there is only one way to do this we determine the P (A and B) = 1/52

So P (A or B) = 1/4+1/13 - 1/52 = 16/52

 Find P (A or C) .

This one is easy since if you select one card from the deck you can note get both a heart and a black card the probability is 0. Events A and C are said to be mutually exclusive since they can not occur simultaneously.

The formal addition rule is nice, yet we do not always have to use it if we remember the intuitive rule above

Here is a table of data:

If one subject is selected at random find the probability of getting

Solution:  The table above is called a contingency table. The first thing you should do is calculate the totals in each column and row - here they are:

So here are the answers:

P(nonsmoker)=350/1000

P(Died from heart disease)=465/1000

P( Nonsmoker who dies from cancer) = 55/190 - not the sample space here contains ONLY nonsmokers so the total is 190

P( Someone who died from heart disease or a person who is a nonsmoker) = P(Someone who died from heart disease )+P(a person who is a nonsmoker)-P(Someone who died from heart disease and a person who is a nonsmoker) = 465/1000+350/1000-155/1000 = 660/1000

P(A smoker or someone who died from cancer)=P(A smoker) + P(someone who died from cancer) - P(smoker and someone who died from cancer) = 650/1000+190/1000-135/1000=705/1000

Sometimes this compound event is easy to calculate - for example in the experiment where one card is drawn from a deck and you want the probability of both a Jack and a red card (What is it?) Or in the table above, probability someone died from cancer and is a nonsmoker

More difficult when A occurs in one trial and B occurs in the next

Here is a tree diagram to illustrate the situation:

Let A = event that you choose a plaid shirt P (A) = 1/3

Let B = event that you choose yellow pants P (B) = 1/2

P (A and B) = 1/6 (see tree diagram)

Notice that this illustrates the rule

Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other,

For example you toss a coin twice - the outcome of the first toss does not affect the outcome of the second.

Not ALWAYS the case of independence - consider this example

EXPERIMENT: Drawing two cards from a standard deck without replacement.

P (A and B) = 1/52

P (A and C) = 13/52 * 13/51

P (B and D) = 4/52 * 4/51

Events A and C (and B and D) are not independent - they are dependent. If you select a heart on the firsat card, it will affect the probability of the event C. In general we can use this rule

P(B | A) is conditional probability notation: The conditional probability of B given A is the probability of event B occurring, given that A has already occurred. We can calculate it by assuming A has occurred and operating under that assumption, calculate the probability that B will occur

P (C I A) = 13/51

P (D | B) = 4/51

If event E is second card is a heart

P (E I A) = 12/51

P (A | B) = 1/4

P (B I A) = 1/13

More difficult examples:

P (smoker | died from cancer) - since we know the person died from cancer, we can reduce our sample space down to 190, the number of smokers in this group is 135, so probability is 135/190

P (Die from other | non-smoker) - the number of non-smokers is 350, Those that died from other causes is 140, so 140/350

A couple decides to have 4 children - what is the probability of at least one of the children being a girl?

Here is a tree diagram:

Note that there are only 5 outcomes

No girls

1 girl

2 girls

3 girls

and

4 girls

The probability of at least one girl is P(1 girl) + P(2 girls) + P(3 girls) + P(4 girls)

But since P(no girls) + P(1 girl) + P(2 girls) + P(3 girls) + P(4 girls) = 1, we can rewrite this as

Doing the problem this way we get:

P(at least one girl) = 1 - P(no girls) = 1 - 1/16 = 15/16

Solution: Since you are testing 10, the batch will be rejected if 1 or more of the light bulbs are defective. This is saying that at least one light bulb is defective, so

P(at least one light bulb is defective) = 1 - P(no light bulb is defective)

= 1 - =0.262576