### 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." ThenP(A or B) is the probability thateitherone die shows a 3orthe 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, thatP(A) = 6/36. Now, consider a slightly different question: what isP(A|B) (that is, what is the probability of the sum being 7given thatone 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. |

- Are A and B independent?
- Are A and C independent?

**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.

**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.

### Ask a Question

Have questions about this article or topic? Ask### Related Questions

#### Q:

#### Q:

#### Q:

#### Q:

### Popular Articles

- Kindergarten Sight Words List
- First Grade Sight Words List
- 10 Fun Activities for Children with Autism
- Signs Your Child Might Have Asperger's Syndrome
- Theories of Learning
- A Teacher's Guide to Differentiating Instruction
- Child Development Theories
- Social Cognitive Theory
- Curriculum Definition
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development