Answer to Question #328665 in Statistics and Probability for Fraign

Question

Box A and Box B both contain the numbers 2, 4, 6 and 8. Construct the probability mass function and draw the histogram of the sum when one number from each box is taken at a time, with replacement.

Answer 1

$Y=X_1+X_2\\P\left( Y=4 \right) =P\left( 2,2 \right) =\frac{1}{4}\cdot \frac{1}{4}=0.0625\\P\left( Y=6 \right) =P\left( 2,4 \right) +P\left( 4,2 \right) =\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}=0.125\\P\left( Y=8 \right) =P\left( 2,6 \right) +P\left( 4,4 \right) +P\left( 6,2 \right) =\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}=0.1875\\P\left( Y=10 \right) =P\left( 2,8 \right) +P\left( 4,6 \right) +P\left( 6,4 \right) +P\left( 8,2 \right) =\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}=0.25\\P\left( Y=12 \right) =P\left( 4,8 \right) +P\left( 6,6 \right) +P\left( 8,4 \right) =\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}=0.1875\\P\left( Y=14 \right) =P\left( 6,8 \right) +P\left( 8,6 \right) =\frac{1}{4}\cdot \frac{1}{4}+\frac{1}{4}\cdot \frac{1}{4}=0.125\\P\left( Y=16 \right) =P\left( 8,8 \right) =\frac{1}{4}\cdot \frac{1}{4}=0.0625$