Heya, could someone please explain how they found the number of terms? I understand the rest of the question
Since we want to find when Alan eventually picks a
black jellybean, we need to consider every case.
So, if Alan picks a black jellybean on his first go, it's simply 1/n
3rd go: (n - 1)/n * (n - 2)/(n - 1) * 1/(n - 2), since the first and second times, we don't want the black jellybean.
5th go: (n - 1)/n * (n - 2)/(n - 1) * (n - 3)/(n - 2) * (n - 4)/(n - 3) * 1/(n - 4), same reasoning.
So, we sum this up to get when Alan eventually picks a black jellybean.
P(Alan picks a black jellybean) = 1/n + (n - 1)/n * (n - 2)/(n - 1) * 1/(n - 2) + (n - 1)/n * (n - 2)/(n - 1) * (n - 3)/(n - 2) * (n - 4)/(n - 3) * 1/(n - 4) + ... + (n - 1)/n * (n - 2)/(n - 1) ... * 2/3 * 1/2 * 1
We can actually rewrite this as:
P(Alan picks a black jellybean) = 1/n + 1/n * (n - 1)/(n - 1) * (n - 2)/(n - 2) + 1/n * (n - 2)/(n - 1) * (n - 2)/(n - 2) * (n - 3)/(n - 3) * (n - 4)/(n - 4) + ... + 1/n * (n - 1)/(n - 1) ... * 3/3 * 2/2 * 1
= 1/n + 1/n + 1/n + ... 1/n
How do we know it's (n + 1)/2 terms?Let's consider each term independently.
Case 1: 1st go -- 1/n
Case 2: 3rd go -- 1/n
Case 3: 5th go -- 1/n
Case 4: 7th go -- 1/n
We can develop a pattern here between the case number and the term we're interested in.
Case n: (2n - 1)th go -- 1/n - This is the general pattern.
Thus, the nth go is:
Case [(n + 1)/2]: nth go -- 1/n
This means that there are (n + 1)/2 terms.
Which means that the probability is (1/n) * (n + 1)/2 = (n + 1)/2n as required.