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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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