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______
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A) .081 A).24
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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
A. at least 6 days means either 6 days or 7 days. So add the probabilities P(6)+P(7) to get the answer.
I got .0208 Is that right ?
How did you get P(6), did you do