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![(n-r)]/r!<br/>(n-r)!. Here n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore we have P(15) = 20!(0.5](https://www.braindumpsqa.com/_/bsqa/imgs/product.jpg)
Exam Code: (n-r)]/r!
(n-r)!. Here n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore we have P(15) = 20!(0.5
Exam Name: 15)(0.5Certification
Version: V16.75
Q & A: 400 Questions and Answers
(n-r)]/r!
(n-r)!. Here n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore we have P(15) = 20!(0.5 Free Demo download
About 5)/15!5! = 0.0148 P(16) = 20!(0.5 (n-r)]/r!
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NEW QUESTION: 1
What is a risk when initiating the containment of a rogue AP?
A. disassociating clients of valid access points that are operated by a neighboring organization
B. breaking the radio of the containing AP
C. breaking the rogue client radio or its firmware
D. disrupting transmission of neighboring AP clients
Answer: A
Explanation:
Explanation/Reference:
Explanation:
NEW QUESTION: 2
The sponsors of a well-known charity came up with a unique idea to attract wealthy patrons to the $500 a plate dinner. After the dinner, it was announced that each patron attending could buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50-50. If you bought a set of
20 tickets, what is the chance that you will win 15 or more prizes?
A. 0.750
B. 0.006
C. None of these answers
D. 0.250
E. 0.021
Answer: E
Explanation:
Explanation/Reference:
Explanation:
This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by: n!(p
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