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

(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

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 3.6.1Venn diagram showing events and .

We can rewrite ( 3.5.1) in the form

(3.6.3)

The rule provided by ( 3.6.3) is known as Bayes's theorem for two events E and F ; the probabilities and are sometimes referred to as the prior probabilities of events F and , respectively (note that ). The conditional probability as given by Bayes's theorem ( 3.6.3), is referred to as the posterior probability of F , given that the event E has occurred. An interpretation of ( 3.6.3) is that, posterior to observing that the event E has occurred, the probability of F changes from , the prior probability, to , the posterior probability.

Example 3.6.1(Bayes's theorem in action) The Gimmick TV model A uses a printed circuit, and the company has a routine method for diagnosing defects in the circuitry when a set fails. Over the years, the experience with this routine diagnostic method yields the following pertinent information: the probability that a set that fails due to printed circuit defects (PCD) is correctly diagnosed as failing because of PCD is 80%. The probability that a set that fails due to causes other than PCD has been diagnosed incorrectly as failing because of PCD is 30%. Experience with printed circuits further shows that about 25% of all model A failures are due to PCD. Find the probability that the model A set's failure is due to PCD, given that it has been diagnosed as being due to PCD.