Question

A roulette wheel is constructed in such a way that the numbers from 1 to 10 can fall into it and the probability of dropping a number (k + 1) is 2 times greater than dropping the number k (for k = 1; 2; 3; ... 9). Find the probability that: a falls the number 1; b to drop the number 10. Give the ending as a decimal number rounded to 3 digits after the decimal point.

Answer 1

$\text{Let }X\text{ be our random variable (the number that falls)}.\\ x\text{ --- probability that the number 1 falls}\\ 2x\text{ --- probability that the number 2 falls}\\ 4x\text{ --- probability that the number 3 falls}\\ x+2x+4x+\ldots=1\\ \text{The left side is the sum of the geometric progression with }n=10,\ q=2,\ b_1=1.\\ S_{n}=\frac{b_1(1-q^n)}{1-q}=\frac{1(1-2^{10})}{1-2}=1023.\\ 1023x=1\\ x=\frac{1}{1023}\approx 0.001.$

a. So the probability that the number 1 falls is $0.001$.

The probability that the number 2 falls is $\frac{2}{1023}\approx 0.002.$

$b_{10}=b_1q^9=1\cdot 2^9=512.$

b. The probability that the number 10 falls is $\frac{512}{1023}\approx 0.500.$