Assume there is a bakery that offers two types of Pie: A and B. There are 10 customers in total who come in everyday and order one of the two pies. The probability that a customer chooses pie A is .5 and pie B is .5.
How many Pies does the bakery need to make sure that every customer gets their choice of pie 95% of the time. I know the answer but have two methods that seem to have came up with the same answer and would like to know which one is correct.
first we need to see how many people will choose pie A. This is given by the binomial distribution.
$${{10}\choose{j}}.5^j.5^{10-j}$$
Method 1:
The first method I have seen took the binomial distribution and just keep restricting the number of pie A's chosen until you were under 95%. For example:
$$ \sum_{j=0}^{10}{{10}\choose{j}}.5^j.5^{10-j} = 1$$ $$ \sum_{j=1}^{9}{{10}\choose{j}}.5^j.5^{10-j} \approx .998$$ $$ \sum_{j=2}^{8}{{10}\choose{j}}.5^j.5^{10-j} \approx .9785$$ $$ \sum_{j=3}^{7}{{10}\choose{j}}.5^j.5^{10-j} \approx .773$$
So here we would say we would need 8 of each type to make sure that our customers can get what they prefer 95% of the time. Because once we look at the probability of choosing between 3 and 7 of the same type of pie our probability has dropped below .95
Method 2:
I also have seen this done using the below.
$$ \sum_{j=10}^{10}{{10}\choose{j}}.5^j.5^{10-j} \approx .00097$$ $$ \sum_{j=9}^{10}{{10}\choose{j}}.5^j.5^{10-j} \approx .0107$$ $$ \sum_{j=8}^{10}{{10}\choose{j}}.5^j.5^{10-j} \approx .0546$$
Here we would say that of our 10 customers there is approximately a 5% chance that 8 or more of them choose the same type of pie. So if we have 8 of each pie on hand our customers will be able to have their choice of pie approximately 95% of the time.
Method 2 makes sense to me but I am interested to know which method is the correct method or if they both are.
They're equivalent; ${n\choose k}={n\choose n-k}$.