# Cereal boxes

• Jan 27th 2011, 12:39 PM
dwsmith
Cereal boxes
A cereal manufacturer is evaluating a new design for the box of its leading brand. Ten people are shown the new and old box design. Each is asked which one he or she likes best. If each person doesn't have a preference and simply selects one of the two designs at random, what is the probability of each of all 10 selecting the new design?

Why is the total number of combination possibilities $2^{10}\text{?}$
• Jan 27th 2011, 12:46 PM
Plato
The are two choices and ten people.
The number of functions from a set of ten to a set of two is $2^{10}$.
• Jan 27th 2011, 12:48 PM
Ithaka
Quote:

Originally Posted by dwsmith
A cereal manufacturer is evaluating a new design for the box of its leading brand. Ten people are shown the new and old box design. Each is asked which one he or she likes best. If each person doesn't have a preference and simply selects one of the two designs at random, what is the probability of each of all 10 selecting the new design?

Why is the total number of combination possibilities $2^{10}\text{?}$

Because each person has 2 choices. For each of the 2 choices person 1 has, person 2 has 2 choices.
So if we work only with 2 persons, the number of choices will be 2 * 2 = $2^2$
If we extend the same reasoning to 3 persons, the number of choices will be 2*2*2
= $2^3$

etc.
• Jan 27th 2011, 12:56 PM
dwsmith
Now, how can I figure out the probability that 9, 8, 7, and 6 people select the new design?
• Jan 27th 2011, 01:00 PM
Plato
Quote:

Originally Posted by dwsmith
Now, how can I figure out the probability that 9, 8, 7, and 6 people select the new design?

This is the number $\sum\limits_{k = 6}^9 {\dbinom{10}{k}}$. WHY?

I should that is the number of 6,7,8, or 9 prefer the new design.
• Jan 27th 2011, 01:01 PM
dwsmith
Quote:

Originally Posted by Plato
This is the number $\sum\limits_{k = 6}^9 {\dbinom{10}{k}}$. WHY?

Because 10 choose 6 + 7 + ... + 9 is the total.