Question

A roulette wheel is constructed in such a way that the numbers can fall from 1 to 100 and the probability of a humerus falling (k + 1) is 2 times greater than a numerical drop (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

$\Sigma_{i=1}^{100}p_i=1,\text{where }p_i \text{probability of getting a number } i$

$\Sigma_{i=1}^{100}p_i=\Sigma_{i=1}^{10}p_i+\Sigma_{i=11}^{100}p_i$

$\Sigma_{i=11}^{100}p_i=0.9$

$\Sigma_{i=1}^{10}p_i=0.1\text{ where }p_i=p_{i-1}*2$

$\Sigma_{i=1}^{10}p_i=\Sigma_{i=1}^{10}p_1*2^{i-1}=p_1*\Sigma_{i=1}^{10}2^{i-1}=1023*p_1$

$p_1=\frac{0.1}{1023}\approx0$

$p_{10}=512*p_1=\frac{512}{1023}*0.1\approx0.05$

Answer: $p_1=0;p_{10} =0.05 \text{ rounded to 3 digits after the decimal point.}$