Practice problems for these concepts can be found at:

- Probability Solved Problems for Beginning Statistics
- Probability Supplementary Problems for Beginning Statistics

### Multiplication Rule for the Intersection Of Events

The *intersection* of two events A and B consists of all those outcomes which are common to both A and B. The intersection of the two events is represented as A *and* B. The intersection of two events is also represented by the symbol A B, read as A *intersect* B. A Venn diagram representation of the intersection of two events is shown in Fig. 4-6.

The probability of the intersection of two events is given by the *multiplication rule*. The multiplication rule is obtained from the formula for conditional probabilities, and is given by formula (*4.9*).

- P(A

*and*B) = P(A) P(B | A) (

*4.9*)

**EXAMPLE 4.22** A small hospital has 40 physicians on staff, of which 5 are cardiologists. The probability that two randomly selected physicians are both cardiologists is determined as follows. Let A be the event that the first selected physician is a cardiologist, and B the event that the second selected physician is a cardiologist. Then P(A) = = .125, P(B | A) = = .103, and P(A *and* B) = .125 × .103 = .013. If two physicians were selected from a group of 40,000 of which 5000 were cardiologists, then P(A) = = .125, P(B | A) = = .125, and P(A *and* B) = (.125)^{2} = .016. Notice that when the selection is from a large group, the probability of selecting a cardiologist on the second selection is approximately the same as selecting one on the first selection.

Following this line of reasoning, suppose it is known that 12.5% of all physicians are cardiologists. If three physicians are selected randomly, the probability that all three are cardiologists equals (.125)^{3} = .002. The probability that none of the three are cardiologists is (.875)^{3} = .670.

If events A and B are independent events, then P(A| B) = P(A) and P(B|A) = P(B). When P(B| A) is replaced by P(B), formula (*4.9*) simplifies to

- P(A

*and*B) = P(A) P(B) (

*4.10*)

**EXAMPLE 4.23** Ten percent of a particular population have hypertension and 40 percent of the same population have a home computer. Assuming that having hypertension and owning a home computer are independent events, the probability that an individual from this population has hypertension and owns a home computer is .10 × .40 = .04. Another way of stating this result is that 4 percent have hypertension and own a home computer.

### Addition Rule for the Union of Events

The *union* of two events A and B consists of all those outcomes that belong to A or B or both A and B. The union of events A and B is represented as A B or simply as A *or* B. A Venn diagram representation of the union of two events is shown in Fig. 4-7. The darker part of the shaded union of the two events corresponds to overlap and corresponds to the outcomes in both A and B.

To find the probability in the union, we add P(A) and P(B). Notice, however, that the darker part indicates that P(A *and* B) gets added twice and must be subtracted out to obtain the correct probability. The resultant equation is called the *addition rule* for probabilities and is given by formula (*4.11*).

- P(A

*or*B) = P(A) + P(B) – P(A

*and*B) (

*4.11*)

If A and B are mutually exclusive events, then P(A *and* B) = 0 and the formula (*4.11*) simplifies to the following:

- P(A

*or*B) = P(A) + P(B)

**EXAMPLE 4.24** Forty percent of the employees at Computec, Inc. have a college degree, 30 percent have been with Computec for at least three years, and 15 percent have both a college degree and have been with the company for at least three years. If A is the event that a randomly selected employee has a college degree and B is the event that a randomly selected employee has been with the company at least three years, then A *or* B is the event that an employee has a college degree or has been with the company at least three years. The probability of A *or* B is .40 + .30 – .15 = .55. Another way of stating the result is that 55 percent of the employees have a college degree or have been with Computec for at least three years.

**EXAMPLE 4.25** A hospital employs 25 medical-surgical nurses, 10 intensive care nurses, 15 emergency room nurses, and 50 floor care nurses. If a nurse is selected at random, the probability that the nurse is a medical-surgical nurse or an emergency room nurse is .25 + .15 = .40. Since the events of being a medical-surgical nurse and an emergency room nurse are mutually exclusive, the probability is simply the sum of probabilities of the two events.

Practice problems for these concepts can be found at:

### Ask a Question

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

See More Questions### 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
- A Teacher's Guide to Differentiating Instruction
- Definitions of Social Studies
- What Makes a School Effective?
- Curriculum Definition
- Theories of Learning
- Child Development Theories