I need some help with the next question in probability:
In the range {1,2,...,100}, someone picks randomly 15 different numbers, with the same probability for each number.
- What is the average sum of the 15 numbers (no duplicates are allowed)?
- Assume that the probability of choosing even number is 3 times bigger than choosing odd number. What will be the average sum now?
1 . $\frac{15}{100}=\frac{3}{20}$ is the probability for each number to be picked. Hence, the expected value of the sum is
$1\cdot \frac{3}{20}+2\cdot \frac{3}{20}+\cdots + 100\cdot \frac{3}{20}= \frac{3}{20} \cdot (1+2+\cdots 100)=757.5$
2 . Let $p$ be the probability for an odd number to be picked. Then the probability for an even number to be picked is $4p$. (There might be confusion here. Mathematically, $3$ times bigger than means $4$ times as big as, at least to my understanding. But in daily life, people who say $3$ times bigger than usually mean $3$ times as big as.) There are $50$ even numbers and $50$ odd numbers. So, $p \cdot 50 + 4p \cdot 50=15$. We obtain $p=\frac{3}{50}, 4p=\frac{12}{50}$. Therefore, the expected value of the sum is
$1\cdot \frac{3}{50} + 3\cdot \frac{3}{50}+\cdots + 99\cdot \frac{3}{50}+2\cdot \frac{12}{50} +4\cdot \frac{12}{50}+\cdots+100\cdot\frac{12}{50}$ $=\frac{3}{50}\cdot(1+3+\cdots99)+\frac{12}{50}\cdot(2+4+\cdots+100)$