How many numbers must be taken from a table of random…
How many numbers must be taken from a table of random numbers so that we can be confident with probability $0.9$ that at least one even number will be among them?
Explanation of Method
In a table of random numbers each number is even with probability $\frac12$, independently of the others. The probability that $n$ numbers contain no even number is $\left(\frac12\right)^n$. We need $1-\left(\frac12\right)^n\ge0.9$, i.e. $\left(\frac12\right)^n\le0.1$, and solve for the smallest integer $n$.
Step 1: One number
The last digit of a random number is equally likely to be any of $0,\dots,9$; five of them are even. So $P(\text{even})=\frac12$ and $P(\text{odd})=\frac12$, independently for different numbers.
Step 2: $n$ numbers
\[ P(\text{at least one even}) = 1-P(\text{all odd}) = 1-\left(\tfrac12\right)^n. \]
Step 3: Solve the inequality
\[ 1-\left(\tfrac12\right)^n\ge0.9 \iff 2^n\ge10 \iff n\ge\log_2 10\approx3.32. \]
\[ \boxed{n=4.} \]
Step 4: Check
| $n$ | all odd | at least one even |
|---|---|---|
| $1$ | $0.5000$ | $0.5000$ |
| $2$ | $0.2500$ | $0.7500$ |
| $3$ | $0.1250$ | $0.8750$ |
| $4$ | $0.0625$ | $0.9375$ ✓ |
| $5$ | $0.0312$ | $0.9688$ |
Three numbers give only $0.875\lt0.9$; four numbers give $0.9375\ge0.9$.
Result
| Answer | |
|---|---|
| Condition | $1-(1/2)^n\ge0.9\iff 2^n\ge10$ |
| Minimum number | $n=4$ (probability $0.9375$) |