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Question 21 According to Internet security experts, approximately 90% of all e-mail messages are spam (unsolicited commercial e-mail), while the remaining 10% are legitimate. A system administrator wishes to see if the same percentages hold true for the e-mail traffic on her servers. She randomly selects e-mail messages and checks to see whether or not each one is legitimate. (Unless otherwise specified, round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.) Assuming that 90% of the messages on these servers are also spam, compute the probability that the first legitimate e-mail she finds is the seventh message she checks. Compute the probability that the first legitimate e-mail she finds is the seventh or eighth message she checks. Compute the probability that the first legitimate e-mail she finds is among the first seven messages she checks. On average, how many messages should she expect to check before she finds a legitimate e-mail? (Round your answer to one decimal place.) messages Submit Question Jump to Answer O Type here to search DI 1 ΕΑ Presen # 3 2 A + % 5 4 & 7. 6 8

Answer 1

21. 1. P(1st legitimate on 7th attempt) P(6 spam, 7th legi) (0.90) (0.1) = 0.0531 2. P(1st leg on 7th or Sth) = P(1st leg on 7th) + P(1st leg on 8th) = (0.9) (0.1) +(0.9) (0.1) = 0.1010 3. P(1st leg in 7) = C1(0.1)(0.9)® = 0.3720 4. E(no.of trials until 1st leg) = 1/0.1 = 10, so E(no.of trials BEFORE 1st leg) = 10 – 1=9

$18.\ a)\ P(7)=^{25}C_{7}0.27^{7}(1-0.27)^{18} = 0.1743 \\ b)\ P(< 7)=P(\le 6) = \sum_{k=0}^{6}^{25}C_{k}(0.27)^{k}(1-0.27)^{25-k}=0.4692 \\ c)\ E(no.of\ trials\ until\ 1st\ success)=1/success\ probty = 1/0.27 = 3.7037\\ d)\ P(>8)=\sum_{k=9}^{50}^{50}C_{k}(0.27)^{k}(0.73)^{50-k}=0.9497 \\ e)\ P(11\le X\le 16)=\sum_{k=11}^{16}^{50}C_{k}(0.27)^{k}(0.73)^{50-k}= 0.661 \\ f)\ P(5\ until\ 1st\ success)=(0.73)^{4}(0.27)=0.0767 \\ g)\ P(more\ than\ 7\ until\ 1st\ success)=1 - P(\le 7\ until\ 1st\ success)=\\ = .1105$

$Q7 : P(3\ R, 6\ B, 6\ G)=\frac{^{6}C_{3}^{10}C_{6}^{11}C_{6}}{^{27}C_{15}} = 0.1116$