Chapter 2 Section 2 Question 8
Question
Suppose that a bowl contains 10 chips, each uniquely numbered 0 through 9. The chips are thoroughly mixed, one is drawn and the number on it, $X_1$, is noted. This chip is then replaced in the bowl. A second chip is drawn and the number on it, $X_2$, is noted. Compute $P(W = w)$ for every real number w when $W = X_1 + 10X_2$.
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 14 2020
The probability of drawing any single chip is $\frac{1}{10}$
There are 100 elements in W, it contains all integers 0-100 inclusive. $X_2$ represents the first digit and $X_1$ the second(left to right). Since all possible outcomes are equally probable. We can generalize this and write:
$P(W=w)=0.1*0.1=0.01$ for all $w\in[0,99]\cap \mathbb{Z}$
Otherwise:
$P(W=w)=0$ for all $w\notin[0,99]\cap \mathbb{Z}$
Below I have listed out the first ten probability's for fun.
Probability that $X_1 + 10X_2=0$: $$P(W=0)=P(X_1=0)*P(X_2=0)$$ $$P(W=0)=0.1*0.1$$ $$P(W=0)=0.01$$ Probability that $X_1 + 10X_2=1$: $$P(W=1)=P(X_1=1)*P(X_2=0)$$ $$P(W=1)=0.1*0.1$$ $$P(W=1)=0.01$$ Probability that $X_1 + 10X_2=2$: $$P(W=2)=P(X_1=2)*P(X_2=0)$$ $$P(W=2)=0.1*0.1$$ $$P(W=2)=0.01$$ Probability that $X_1 + 10X_2=3$: $$P(W=3)=P(X_1=3)*P(X_2=0)$$ $$P(W=3)=0.1*0.1$$ $$P(W=3)=0.01$$ Probability that $X_1 + 10X_2=4$: $$P(W=4)=P(X_1=4)*P(X_2=0)$$ $$P(W=4)=0.1*0.1$$ $$P(W=4)=0.01$$ Probability that $X_1 + 10X_2=5$: $$P(W=5)=P(X_1=5)*P(X_2=0)$$ $$P(W=5)=0.1*0.1$$ $$P(W=5)=0.01$$ Probability that $X_1 + 10X_2=6$: $$P(W=6)=P(X_1=6)*P(X_2=0)$$ $$P(W=6)=0.1*0.1$$ $$P(W=6)=0.01$$ Probability that $X_1 + 10X_2=7$: $$P(W=7)=P(X_1=7)*P(X_2=0)$$ $$P(W=7)=0.1*0.1$$ $$P(W=7)=0.01$$ Probability that $X_1 + 10X_2=8$: $$P(W=8)=P(X_1=8)*P(X_2=0)$$ $$P(W=8)=0.1*0.1$$ $$P(W=8)=0.01$$ Probability that $X_1 + 10X_2=9$: $$P(W=9)=P(X_1=9)*P(X_2=0)$$ $$P(W=9)=0.1*0.1$$ $$P(W=9)=0.01$$