Engr 323 Homework Set 2 Page 1 of 6 Porteous 227


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1 Engr 323 Homework et 2 Page 1 of 6 The three most popular options on a certain type of new car are automatic transmission (), power steering (), and a radio (). If 70% of all purchasers request, 80% request, 75% request, 85% request or, 90%request or, 95% request or, and 98% request or or, compute the probabilities of the following events. a. The next purchaser will select at least one of the three options ( ). b. The next purchaser will select none of the three options. c. The next purchaser will select only a radio. d. The next purchaser will select exactly one of the three options. The symbolic representation of the given information: P() = 0.70 P() = 0.80 P() = 0.75 P( ) = 0.85 P( ) = 0.90 P( ) = 0.95 P( ) = 0.98 a. The probability that the next purchaser will select at least one of the three options is determined with the following equations (Devore, Page 54) Where: P( ) = P() + P() + P() P( ) P( ) P( ) + P( ) Equation 1 P( ) = P() + P()  P( ) Equation 2 P( ) = P() + P() P( ) Equation 3 P( ) = P() + P() P( ) Equation 4 P( ) = P() P() P() + P( ) + P( ) + P( ) + P( ) Equation 5
2 Engr 323 Homework et 2 Page 2 of 6 elow I will explain why Equation 1 is true for any events, and. The equations can be demonstrated pictorially with Venn diagrams. The total area in the Venn diagram represents the sample space (). The three circular areas represent the probability of each option being selected. The probability that the next purchaser will select one or more of the three options is P()+P()+P() if and only if the 3 options are disjoint. This probability is illustrated by the blue area in Figure 1. P( ) = P() + P() + P() Figure 1. P( ) = P() + P() + P() for disjoint events However, the probability that the purchaser will select at least one option is not accurately represented by simply the sum of the areas of the three circles as shown in Figure 1 because a purchaser might choose more than one option so they are not disjoint events Figure 2 is a more accurate diagram of how, and can be depicted. In this case, when considering P( )=P()+P()+P(). The areas of the overlapping circles are counted twice so one of the areas common to two circles (the football shaped area) is subtracted from the total. The blue portion of Figure 2 represents one of the three football shaped areas that is counted twice. Figure 2. The area represented is counted twice when considering P()+P()=P()=P( ) for nondisjoint events.
3 Engr 323 Homework et 2 Page 3 of 6 ince all three areas common to two options are subtracted from the total, the area common to all three options is subtracted also and must be accounted for. The final correction is to add this common area back to the sum of all three options. This portion of the equation is explained in the text on page 55. Figure 3. P( ) Thus P( )=P()+P()=P()P( )P( )P( )+P( ) To find the probability that the next purchaser will select at least one option, Equations 2, 3, 4 and 5 will be solved and plugged into Equation 1. P( ) = P( ) = 0.65 P( ) = P( ) = 0.55 P( ) = P( ) = 0.60 P( ) = P( ) = 0.53 P( ) = P( ) = 0.98 The probability that the next purchaser will select at least one option is This probability is denoted in Figure 4.
4 Engr 323 Homework et 2 Page 4 of Figure 4. Probability that the next purchaser will select at least one option is b. The probability that the next purchaser will select none of the three options requires analysis of complement events. If two events are complement, the probability of an outcome in either event is 1. The probability that an event does not occur is then simply one minus the probability that the event will occur. The blue area of Figure 5 represents the probability that no options are selected. 1 = P() = P( ) (Devore Page 52) 1 = P() + P( ) P( ) = 1  P() ince P( ) is known, P[( ) ] = P(none selected) can be determined. P[( ) ] = 1 P( ) Equation 6 The probability that the next purchaser selects none of the three options is: (2) P[( ) ] = P[( ) ] = Figure 5. The probability that no options are selected is 0.02 c. The probability that the next purchaser will select only a radio () is determined by subtracting the probability that the purchaser will select either a radio () and an automatic transmission () or a
5 Engr 323 Homework et 2 Page 5 of 6 radio () and power steering () from the probability that the next purchaser will select a radio in any configuration. Remember, the probability of selecting all three options must be added back to the sum because it was removed in the last step as shown in Equation 7. The blue area of Figure 6 illustrates the probability that only a radio will be selected by the next purchaser. P(radio only) = P( ) = P() P( ) P( ) + P( ) Equation 7 The probability that only a radio will be selected is P(radio only) = P(radio only) = Figure 6. The probability of only a radio selected is.13 d. The probability that the next purchaser will select only one of the options is determine similarly to the probability that any one specific option will be selected alone (as in part c). The probability that only one of the options will be selected is simply the sum of the probabilities that each option is selected alone. These probabilities can be added because they are disjoint (as explained on page 53 of the textbook). Equation 8 shows the set theory notation of the probability of only one option selected. The blue area of Figure 7 represents the probability that only one option is selected. P(only one option) = P[( ) ( ) ( )] Equation 8 s explained in part c. P( ) = P() P( ) P( ) + P( ) imilarly: P( ) = P() P( ) P( ) + P( ) and, P( ) = P() P( ) P( ) + P( ) P(only one option) = P() P( ) P( ) + P( ) + P() P( ) P( ) + P( ) + P() P( ) P( ) + P( )
6 Engr 323 Homework et 2 Page 6 of 6 The probability that the next purchaser will select only one option is P(only one option) = P(only one option) = 0.24 We may choose to use Figure 4 to determine the probability the next purchaser will select only one option instead of analyzing Equation 8. Figure 4 shows the probability that each option is selected alone. ince we know they are disjoint, we can sum the probabilities. P(only one option)= = Figure 6. The probability of only one option selected is =.24
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