Question

suppose you roll two fair, 6 - sided dice - one red and one blue. are the events sum is 7 and blue die shows a 4 independent? justify your answer.

$p$(sum is $7 |$ blue die is a $4$) $=$
$p$(sum is $7 |$ blue die is not a $4$) $=$
the probabilities so the events blue die shows a 4 andsum is 7 independent.
knowing that a roll has a blue die showing a 4 the probability that the sum is 7.

Explanation:

Step1: Calculate \(P(\text{sum is }7|\text{blue die is }4)\)

When the blue die shows \(4\), for the sum of the two - dice to be \(7\), the red die must show \(7 - 4=3\).
Since the blue die is fixed at \(4\) (a single outcome for the blue die), and the red die has \(6\) possible outcomes (\(1,2,\cdots,6\)), \(P(\text{sum is }7|\text{blue die is }4)=\frac{1}{6}\).

Step2: Calculate \(P(\text{sum is }7|\text{blue die is not }4)\)

The total number of outcomes when rolling two dice is \(n(S)=6\times6 = 36\).
The pairs \((x,y)\) such that \(x + y=7\) are \((1,6)\), \((2,5)\), \((3,4)\), \((4,3)\), \((5,2)\), \((6,1)\), so \(n(\text{sum is }7)=6\).
The number of outcomes where the blue die is not \(4\): \(n(\text{blue die is not }4)=6\times5 = 30\).
The number of outcomes where the sum is \(7\) and the blue die is not \(4\): There are \(5\) such pairs (\((1,6)\), \((2,5)\), \((4,3)\), \((5,2)\), \((6,1)\)).
Using the formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\), \(P(\text{sum is }7|\text{blue die is not }4)=\frac{5}{30}=\frac{1}{6}\).

Answer:

\(P(\text{sum is }7|\text{blue die is }4)=\frac{1}{6}\), \(P(\text{sum is }7|\text{blue die is not }4)=\frac{1}{6}\). The probabilities are equal, so the events “blue die shows a \(4\)” and “sum is \(7\)” are independent. Knowing that a roll has a blue die showing a \(4\) does not change the probability that the sum is \(7\).