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anonymous
 one year ago
The probability of Andy and Anna having lunch together is 40%. The probability of them having lunch together at least six days a week is _______ and the probability of having lunch exactly 6 times in a week is______
1st Blank 2nd Blank
A) .081 A).24
B).091 B).083
C).002 C).017
anonymous
 one year ago
The probability of Andy and Anna having lunch together is 40%. The probability of them having lunch together at least six days a week is _______ and the probability of having lunch exactly 6 times in a week is______ 1st Blank 2nd Blank A) .081 A).24 B).091 B).083 C).002 C).017

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mathmate
 one year ago
Best ResponseYou've already chosen the best response.01. Assume lunching together each day is independent of whether they ate together the previous days or the next days. 2. p=probability of having lunch together in a particular day. 3. p does not change over the entire week 4. each trial (day) is a Bernoulli trial (i.e. either eat together or not eat together) then we can apply Binomial Probability with parameters p=0.4, n=7 (days). Let k=exactly number of days they lunched together P(r)=C(n,k)p^k (1p)^(nk) where C(n,k)=n!/(k!(nk)!) B. lunched together exactly 7 days, set k=6, C(n,k)=7!/(7!0!)=1 P(7)=1(0.4)^7(0.6)^0=0.0016384 A. at least 6 days means either 6 days or 7 days. So add the probabilities P(6)+P(7) to get the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got .0208 Is that right ?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0How did you get P(6), did you do C(7,6)*(0.4^6)*(0.6^1) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well would you use both the 7 and the 6 in C(7,6) or would it be (7)*(0.4^6)*(0.6^1) and then (6)*(0.4^7)*(0.7^1)?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Do not forget the term C(n,k), and recall C(7,7)=7!/(7!0!)=1, and C(7,6)=7!/(6!1!)=7 because C(n,k)=n!/(k!(nk)!)
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