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1) Once a Woman won $1 Million in scratch off game from a lottery. Some years later, she won $1 million in another scratch off game. In the first game, she beat odds of 1 in 5.2 million to win. In the second, she beat odds of 1 in 805,600. a) What is the probibility that an individual would win 1 million in both games if they bought one scratch off ticket from each game? b) What is the probability that an individual would win $1 million twice in the second scratch of game? Use Scientific Notation. Round to the nearest tenth as needed. Answers: a) (1/5200000)(1/805600) = 2.4 x 10-13 b) (1/805600)2 = 1.5 x 10-12 2) A bionomial probability experiment is conducted with the given parameters. Compuete the probability of x successes in the n independent trials of the experiment. n=9 p=.6 x greater/equal3 Round to four decimal places as needed. Answer: P(x≥3) = 0.9750 3) The number of hits to a website follows a poisson process. Hits occur at the rate of 3.3 per minute between 7 PM and 11 PM. Given below are three scenarios for numbers of hits to the site. Compute the probability of each scenario between 7:26 PM and 7:30 PM. a) Exactly four b)Fewer than four. c) at least four. Answers: Between 7:26 and 7:30 the number of hits has a poisson distribution(X) with = 4*3.3 = 13.2 a) P(x=4)= 0.0023 b) P(x<4)= P(x3)= 0.0009 c)P(x≥4)= 1-P(x3)= 0.0091 4) Suppose that a sample of peanut butter contains .3 insect fragments per gram. Compute the probability that the number of insect fragments in a 4-gram sample of peanut butter is each of the following values or range of values. a)exactly four b)fewer than four c)at least four d)at least one Answers: the number of insect fragments in a 4-gram sample of peanut butter has a poisson distribution(X) with = 4*0.3 = 1.2 a) P(x=4)= 0.0260 b) P(x<4)= P(x3)= 0.9662 c)P(x≥4)= 1-P(x3)= 0.0338 d) P(x≥1)= 1-P(x=0) =0.6988 5) Suppose a life insurance company sells a $270,000 one year term life insurance polict to a 24 year old female for $220. The probability that the female survives the year is .999593. Compute and interpret the expected value of this policy to the insurance company. The expected value is $ ? Answer: The expected value is: 270,000(1-0.999593) = $109.89 On average the company will lose $109.89 per each 24 year old female so the expected earning per person is: $220 - $109.89 = $110.11 6) The random variable X follows a poisson process with the given value of ^ and t Assuming ^=.14 and t =10 compute the following. X has a poisson distribution with parameter = 0.14(10) = 1.4 a) P(5) = P(X=5) = 0.0111 b) P(E<5) = P(X4)=0.9857 c)P(X 5)= 0.9968 d) P(3 X 5) = 0.1633 e) x = x = 1.4