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?
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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^r)[q^(n-r)]/r!(n-r)!. 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!(0.6^10)(0.4^5)/10!(15-10)! = 0.1859
P(11 successes) = 15!(0.6^11)(0.4^4)/11!(15-11)! = 0.1268
P(12 successes) = 15!(0.6^12)(0.4^3)/12!(15-12)! = 0.0634
P(13 successes) = 15!(0.6^13)(0.4^2)/13!(15-13)! = 0.0219
P(14 successes) = 15!(0.6^14)(0.4^1)/14!(15-14)! = 0.0047
P(15 successes) = 15!(0.6^15)(0.4^0)/15!(15-15)! = 0.00047
The sum of all the probabilities is 0.403.