Probability & Statistics Ch.2 Section 2 Question 8

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

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 $W = X_1 + 10X_2$ .

Solution

The probability of drawing any single chip is $\frac{1}{10}$ .

There are 100 elements in W, containing all integers from 0 to 99 inclusive. $X_2$ represents the tens digit and $X_1$ represents the units digit. Since all possible outcomes are equally probable, we can generalize:

P(W=w)=0.1 \times 0.1 = 0.01 \text{ for all } w \in [0,99] \cap \mathbb{Z}

Otherwise: $P(W=w)=0$ for all $w\notin [0,99] \cap \mathbb{Z}$

Below are the first ten probabilities listed out for illustration:

P(W=0)=P(X_1=0) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=1)=P(X_1=1) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=2)=P(X_1=2) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=3)=P(X_1=3) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=4)=P(X_1=4) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=5)=P(X_1=5) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=6)=P(X_1=6) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=7)=P(X_1=7) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=8)=P(X_1=8) \times P(X_2=0) = 0.1 \times 0.1 = 0.01

P(W=9)=P(X_1=9) \times P(X_2=0) = 0.1 \times 0.1 = 0.01