Question

You work at an ice cream stand that offers 6 different flavors. On a busy day you take orders for one scoop cones from 5 people, but forget the flavors. How many flavors do you have to remember for the probability that you randomly get all the other flavors correct on the first try to be greater than 0.01? Explain

Answer 1

it might be to 17
i need to check more aroud my solve equation

Answer 2

Well, this is the say i see it (might be wrong)
If you dont remember any of them, then the probability of you being right is:
\[\frac{1}{6^5}\]
.0001

If you remember one, then its \[\frac{1}{6^4} = .0007\]
Remembering 2 gives:
\[\frac{1}{6^3} = .004\]
Remembering 3 gives:
\[\frac{1}{6^2} = .002\]

So you have to remember 3 flavors.

Answer 3

im sry that should be .02, not .002

Answer 4

You are basically ignoring the people whose order you know, and you want to probability of guessing the rest. If you know k out of the 5 peoples orders, you dont know 5-k of the orders.
Each one of those people had six flavors to choose from, so there are:
6*6*...*6 (5-k times)

\[6^{5-k}\]
possibilities, only one of which is right.
So your probability of being right is:
\[\frac{1}{6^{5-k}} = 6^{-(5-k)} = 6^{k-5}\]
and you want to know when is that greater than .01

Answer 5

so whats the answer?

Answer 6

read my first post =/