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

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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

(3.5.7)

      provided that and are both not zero. Formula (3.5.7) extends to any finite number of events . Note that equation (3.5.5) also can be extended to the case of several mutually independent events so that for these n independent events

(3.5.8)

      Example 3.5.3 (Rolling a die n times) If a true die is thrown n times, what is the probability of never getting an ace (one‐spot)?

      Solution: Let be the event of not getting an ace on the first throw, the event of not getting an ace on the second throw, and so on. Assuming independence of the events and a “true” die, we have . Hence, the required probability from (3.5.8) is

3.6 Bayes's Theorem

      An interesting version of the conditional probability formula (3.5.1) comes from the work of the Reverend Thomas Bayes. Bayes's result was published posthumously in 1763.

      Suppose that E and F are two events in a sample space S and such that . From the Venn diagram in Figure 3.6.1, we can see that the events and are disjoint and that their union is E, so that

(3.6.1)

      Using the rule given by (3.5.6), we can rewrite equation (3.6.1) as

      (3.6.2)

Figure 3.6.1 Venn diagram showing events

and

.

      We can rewrite (3.5.1) in the form

(3.6.3)

      The

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