Bhisham C. Gupta - Statistics and Probability with Applications for Engineers and Scientists Using MINITAB, R and JMP

1 If the promoted technician is a woman, then what is the probability that she is a nongraduate?

2 Find the probability that the promoted technician is a nongraduate when it is not known that the promoted technician is a woman.

Solution:Let S be the sample space associated with this problem, and let E and F be the two events defined as follows:

E: the promoted technician is a nongraduate

F: the promoted technician is a woman

In Part (a) we are interested in finding the conditional probability .

Since any of the 100 technicians could be promoted, the sample space S consists of 100 equally likely sample points. The sample points that are favorable to the event E are 65, and those that are favorable to the event F are 44. Also the sample points favorable to both the events E and F are all the women who are nongraduates and equal to 29. To describe this situation, we have

(a) Therefore,

and for part (b), we have that .

Table 3.5.1 Classification of technicians by qualification and gender.

Note that the probability P ( E ), sometimes known as the absolute probability of E , is different from the conditional probability P(E|F) . If the conditional probability P(E|F) is the same as the absolute probability P(E) , that is, P(E|F) = P(E) , then the two events E and F are said to be independent . In this example, the events E and F are not independent.

Definition 3.5.1Let S be a sample space, and let E and F be any two events in S . The events E and F are called independent if and only if any one of the following is true:

(3.5.3)

(3.5.4)

(3.5.5)

The conditions in equations ( 3.5.3)–( 3.5.5) are equivalent in the sense that if one is true, then the other two are true. We now have the following theorem, which gives rise to the so‐called multiplication rule .

Theorem 3.5.1(Rule of multiplication of probabilities) If E and F are events in a sample space S such that , then

(3.5.6)

Equation ( 3.5.6) provides a two‐step rule for determining the probability of the occurrence of by first determining the probability of E and then multiplying by the conditional probability of F given E .

Example 3.5.2(Applying probability in testing quality) Two of the light bulbs in a box of six have broken filaments. If the bulbs are tested at random, one at a time, what is the probability that the second defective bulb is found when the third bulb is tested?

Solution:Let E be the event of getting one good and one defective bulb in the first two bulbs tested, and let F be the event of getting a defective bulb on drawing the third bulb. Then, is the event whose probability we are seeking. Hence,

The sample space S in which E lies consists of all possible selections of two bulbs out of six, the number of elements in S being . The event E consists of all selections of one good and one defective bulb out of four good and two defective bulbs, and the number of such selections is . Therefore, . We now compute , the probability that F occurs given that E occurs. If E has occurred, there are three good and one defective bulb left in the box. F is the event of drawing the defective one on the next draw, that is, from a box of four that has three good bulbs and one defective. Thus, . The required probability is

For the case of three events, , the extended form of ( 3.5.6) is