A relatively simple problem is stumping me on this 🙁
Given a set of 4 distinct numbers that can be chosen from [0-9], what is the probability that after selecting, the first two numbers would be in ascending order?
Here's my thinking so far:
- If we select 0 as our first, we have a 9/10 chance of choosing an appropriate second number
- If we select 1 as our first, we have an 8/10 chance of choosing an appropriate second number
- …
- If we select 8 as our first, we only have a 1/10 chance of choosing an appropriate second number
- However if we select 9 as our first, we can't select anything greater
So the probability of selecting the first two numbers such that they are ascending should be:
P(1st selection) * P(2nd selection > 1st selection) * P(3rd selection that we don't care about) * P(4th selection that we don't care about) = 1/10 * 1/9 * 1/8 * 1/7 => 1/5040
I don't think I'm going about this problem the right way and my answer is completely off – I don't know how to handle what the probability of selecting the second number would be, since we could choose a great number like 0 for the first, or a bad one like 8 or 9 that would give us less choices. Could someone explain this in a better way?
Best Answer
By symmetry, the probability is $$\frac12.$$ Your approach yields $$ \frac1{10}\cdot\frac99+\frac1{10}\cdot\frac89+\frac1{10}\cdot\frac79+\cdots+\frac1{10}\cdot\frac19=\frac1{10}\cdot\frac19\cdot\frac{9\cdot10}2=\frac12. $$