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- anonymous

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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- anonymous

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- mathmate

1. 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 (1-p)^(n-k) where C(n,k)=n!/(k!(n-k)!)
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

I got .0208 Is that right ?

- mathmate

How did you get P(6), did you do
C(7,6)*(0.4^6)*(0.6^1)
?

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- anonymous

Well 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

Do 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!(n-k)!)

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