P(11 successes) = 15!(0.6 > MCSA: 11)(0.4 > (nr)]/r!
(nr)!. Therefore, we need to find out the probability of getting 10, 11,12,13,14,15 successes and add them up. Here n=15, p=0.6 and q=0.4. r changes from 10 to 15.
P(10 successes) = 15!(0.6
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(nr)]/r!
(nr)!. Therefore, we need to find out the probability of getting 10, 11,12,13,14,15 successes and add them up. Here n=15, p=0.6 and q=0.4. r changes from 10 to 15.
P(10 successes) = 15!(0.6 PDF PackageReal 5)/10!(1510)! = 0.1859
P(11 successes) = 15!(0.6 MCSA: 11)(0.4 (nr)]/r!
(nr)!. Therefore, we need to find out the probability of getting 10, 11,12,13,14,15 successes and add them up. Here n=15, p=0.6 and q=0.4. r changes from 10 to 15.
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(nr)!. Therefore, we need to find out the probability of getting 10, 11,12,13,14,15 successes and add them up. Here n=15, p=0.6 and q=0.4. r changes from 10 to 15.
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P(11 successes) = 15!(0.6 (nr)]/r!
(nr)!. Therefore, we need to find out the probability of getting 10, 11,12,13,14,15 successes and add them up. Here n=15, p=0.6 and q=0.4. r changes from 10 to 15.
P(10 successes) = 15!(0.6 MCSA: 11)(0.4 questions into Topics and Objectives. Real (nr)]/r!
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NEW QUESTION: 1
Sixty percent of the customers of a fast food chain order the Whopper, fries and a drink. If a random sample of 15 cash register receipts is selected, what is the probability that 10 or more will show that the above three food items were ordered?
A. None of these answers
B. 0.403
C. 0.186
D. 1,000
E. 0.000
Answer: B
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 = 1p) is given by: n!(p
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