There is a 0.39 probability that John will purchase a new car,a 0.73 probability
ID: 2918861 • Letter: T
Question
There is a 0.39 probability that John will purchase a new car,a 0.73 probability that Mary will purchse a new car, and a 0.36probability that both will purchase a new car. Find the probabilitythat neither will purchase a new car. ______________ An automobile license plate consists of 3 letters followed by4 digits. How many different plates can be made if repetitions areallowed? If repetitions are not allowed? If repetitions are allowedin the letters but not in the digits? _______________ What is the probability of randomly selecting3 mathematics books and 4 physics books from a box containing5 mathematics books and 9 physics books? ______________ No need to explain it, just show math please. There is a 0.39 probability that John will purchase a new car,a 0.73 probability that Mary will purchse a new car, and a 0.36probability that both will purchase a new car. Find the probabilitythat neither will purchase a new car. ______________ An automobile license plate consists of 3 letters followed by4 digits. How many different plates can be made if repetitions areallowed? If repetitions are not allowed? If repetitions are allowedin the letters but not in the digits? _______________ What is the probability of randomly selecting3 mathematics books and 4 physics books from a box containing5 mathematics books and 9 physics books? ______________ No need to explain it, just show math please.Explanation / Answer
1. There is a 0.39 probability that John will purchase a newcar, a 0.73 probability that Mary will purchse a new car, and a0.36 probability that both will purchase a new car. Find theprobability that neither will purchase a new car.This problem is a "given" problem. Thus, let's state thefacts.
A. P(John) = 0.39
B. P(Mary) = 0.73
C. P(John and Mary) = 0.36
D. P(John or Mary) = P(John) +P(Mary) - P(John and Mary)
E. 1 - Answer from D
F. Finish this and it will beyour answer!
2. An automobile license plate consists of 3 letters followedby 4 digits. a). How many different plates can be made if repetitionsare allowed? You have to do this as a counting rule. There are26 letters in the alphabet and 10 digits (0-9) for the licenseplate. So, you do it like this... 26 * 26 * 26 * 10 * 10 * 10 * 10 = finish this and itwill be your answer! b). If repetitions are not allowed? You have to do this as a counting rule. There are 26letters in the alphabet and 10 digits (0-9) for the licenseplate. So, you do it like this... 26 * 25 * 24 * 10 * 9 * 8 * 7 = finishthis and it will be your answer! So, you do it like this... 26 * 25 * 24 * 10 * 9 * 8 * 7 = finishthis and it will be your answer! c). If repetitions are allowed in the letters but not inthe digits? You have to do this as a counting rule. There are 26letters in the alphabet and 10 digits (0-9) for the licenseplate. So, you do it like this... 26 * 26 * 26 * 10 * 9 * 8 * 7 = finishthis and it will be your answer! So, you do it like this... 26 * 26 * 26 * 10 * 9 * 8 * 7 = finishthis and it will be your answer! 3. What is the probability of randomly selecting3 mathematics books and 4 physics books from a box containing5 mathematics books and 9 physics books? You have to do this as a combination. Formula for thisis: nCr = n! / [(n - r)! r!] 5C3 + 9C4 = finish this and it will be youranswer!
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