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# Probability for AP Statistics (page 2)

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By — McGraw-Hill Professional
Updated on Feb 3, 2011

### Probabilities of Combined Events

P(A or B): The probability that either event A or event B occurs. (They can both occur, but only one needs to occur.) Using set notation, P(A or B) can be written P(A B). A B is spoken as, "A union B."

P(A and B): The probability that both event A and event B occur. Using set notation, P(A and B) can be written P(A ∩ B). A ∩ B is spoken as, "A intersection B."

example: Roll two dice and consider the sum (see table). Let A = "one die shows a 3," B = "the sum is greater than 4." Then P(A or B) is the probability that either one die shows a 3 or the sum is greater than 4. Of the 36 possible outcomes in the sample space, there are 32 possible outcomes that are successes [30 outcomes greater than 4 as well as (1,3) and (3,1)], so

There are nine ways in which a sum has one die showing a 3 and has a sum greater than 4: [(3,2), (3,3), (3,4), (3,5), (3,6), (2,3), (4,3), (5,3), (6,3)], so

Complement of an event A: events in the sample space that are not in event A. The complement of an event A is symbolized by , or Furthermore, P() = 1 – P(A).

### Mutually Exclusive Events

Mutually exclusive (disjoint) events: Two events are said to be mutually exclusive (some texts refer to mutually exclusive events as disjoint) if and only if they have no outcomes in common. That is, A ∩ B = Ø. If A and B are mutually exclusive, then P(A and B) = P(A ∩ B) = 0.

example: in the two-dice rolling experiment, A = "face shows a 1" and B = "sum of the two dice is 8" are mutually exclusive because there is no way to get a sum of 8 if one die shows a 1. That is, events A and B cannot both occur.

### Conditional Probability

Conditional Probability: "The probability of A given B" assumes we have knowledge of an event B having occurred before we compute the probability of event A. This is symbolized by P(A|B). Also,

Although this formula will work, it's often easier to think of a condition as reducing, in some fashion, the original sample space. The following example illustrates this "shrinking sample space."

example: Once again consider the possible sums on the roll of two dice. Let A = "the sum is 7," B = "one die shows a 5." We note, by counting outcomes in the table, that P(A) = 6/36. Now, consider a slightly different question: what is P(A|B) (that is, what is the probability of the sum being 7 given that one die shows a 5)?
solution: Look again at the table:

The condition has effectively reduced the sample space from 36 outcomes to only 11 (you do not count the "10" twice). Of those, two are 7s. Thus, the P(the sum is 7 |one die shows a 5) = 2/11.

alternate solution: If you insist on using the formula for conditional probability, we note that P(A and B) = P(the sum is 7 and one die shows a 5) = 2/36, and P(B) = P(one die shows a 5) = 11/36. By formula

Some conditional probability problems can be solved by using a tree diagram. A tree diagram is a schematic way of looking at all possible outcomes.

example: Suppose a computer company has manufacturing plants in three states. 50% of its computers are manufactured in California, and 85% of these are desktops; 30% of computers are manufactured in Washington, and 40% of these are laptops; and 20% of computers are manufactured in Oregon, and 40% of these are desktops. All computers are first shipped to a distribution site in Nebraska before being sent out to stores. If you picked a computer at random from the Nebraska distribution center, what is the probability that it is a laptop?
solution:

Note that the final probabilities add to 1 so we know we have considered all possible outcomes. Now, P(laptop)

= 0.075 + 0.12 + 0.12 = 0.315

.

### Independent Events

Independent Events: Events A and B are said to be independent if and only if P(A) = P(A|B) or P(B) = P(B|A). That is, A and B are independent if the knowledge of one event | having occurred | does not change the probability that the other event occurs.

example: Consider drawing one card from a standard deck of 52 playing cards.
 Let A = "the card drawn is an ace." P(A) = 4/52 = 1/13. Let B = "the card drawn is a 10, J, Q, K, or A." P(B) = 20/52 = 5/13. Let C = "the card drawn is a diamond." P(C) = 13/52 = 1/4.
1. Are A and B independent?
2. solution: P(A|B) = P(the card drawn is an ace | the card is a 10, J, Q, K, or A) = 4/20 = 1/5 (there are 20 cards to consider, 4 of which are aces). Since P(A) = 1/13, knowledge of B has changed what we know about A. That is, in this case, P(A) ≠ P(A|B), so events A and B are not independent.

3. Are A and C independent?
4. solution: P(A|C) = P(the card drawn is an ace | the card drawn is a diamond) = 1/13 (there are 13 diamonds, one of which is an ace). So, in this case, P(A) = P(A|C), so that the events "the card drawn is an ace" and "the card drawn is a diamond" are independent.

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