Five friends agree to meet at Joe's Coffeehouse in Capital City. However, there are $5$ different Joe's Coffeehouse locations in Capital City, and the friends neglected to agree on which one to meet at, so they each choose one at random. What is the probability that
(a) they all end up at the same Joe's?
(b) they all end up at different Joe's?
For each of those parts I solved it in two different ways and got different answers, I'm not sure which one is correct in both cases. Any explanation as to which is correct, which is wrong, and why would be very well appreciated.
(a) APPROACH 1: Let's fix a friend, there's a ${1\over5}$ chance of him going to each particular Joe's, and then for each of those $5$ options for the remaining $4$ friends, there's a ${1\over5}$ chance of going to that same Joe's, so the probability they all end up at the same Joe's is:$$\left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 = \left({1\over5}\right)^4 = {1\over{625}}$$
APPROACH 2: The probability is going to be$${{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5 \text{ such that }x_i = 5 \text{ for some }i \text{ between }1\text{ and }5 \text{ inclusive}}\over{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5}} = {5\over{\binom{9}{5}}} = {5\over{126}}$$
(b) APPROACH 1: Let's fix a friend, there's a ${1\over5}$ chance of him going to each particular Joe's, and so for that particular Joe's, there's a ${4\over5}$ chance of the next guy going to a different Joe's, a ${3\over5}$ for the next guy etc. so the probability they all end up at different Joe's is:$$5 \left({1\over5}\right)\left({4\over5}\right)\left({3\over5}\right)\left({2\over5}\right) \left({1\over5}\right) = {{24}\over{625}}$$
APPROACH 2: The probability is going to be$${{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5 \text{ such that }x_i = 1 \text{ for all }i \text{ between }1\text{ and }5 \text{ inclusive}}\over{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5}} = {1\over{\binom{9}{5}}} = {1\over{126}}$$
Best Answer
I am unfamiliar with your methods, but I have worked out a solution to both, two different ways, with answers that agree with your answers in approach one.
Method one:
Take the first case:
One friend will definitely go to one shop. The next friend has a $1/5$ probability of visiting that shop, as do the rest of the friends. The probability here is $(1/5)^4=1/625$.
Second case:
One friend will definitely go to one shop. The next friend has a $4/5$ probability of not visiting their shop. The next has $3/5$,... and so on; the total probability is $4!/5^4=24/625$.
Method two:
There are $5\times5\times5\dots=5^5$ total outcomes. Label shops 1-5 as letters A-E. E.g., such an outcome might be “AABED” which says which friend went to which shop.
First case:
Now the event where they are all in one shop happens five times: “AAAAA,BBBBB,...” and so we have probability $5/(5^5)=1/5^4=1/625$, which is what I calculated before.
Second case:
To be all in different shops, we have a string “ABCDE” and all its $5!$ permutations. $5!/5^5=(5\cdot 4!)/5^5=4!/5^4=24/625$ as I said before.
Hope this helps.