Question

A black and a red die are rolled simultaneously. The probability of obtaining a sum greater than 9, given that the black die resulted in a 5 is

• A. 1/2
• B. 1
• C. 2/3
• D. 1/3

Hint:

For "given that" problems involving dice, cards, or coins, it's often easiest to work with the reduced sample space. Simply list all the outcomes that are possible given the condition, and then count how many of those satisfy the event you're interested in.

Answer 1

The Correct Option is D

Step 1: Conceptualization: This scenario involves conditional probability. Given information (the black die's outcome) narrows the sample space, requiring the calculation of another event's probability within this restricted space.

Step 2: Methodology: Define event A as "the sum exceeding 9" and event B as "the black die showing a 5". The objective is to compute P(A|B), the probability of A given B. The formula is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. A simpler approach is to directly analyze the reduced sample space.

Step 3: Detailed Analysis: Employing the reduced sample space method, consider event B, "the black die resulted in a 5", as having occurred. The possible outcomes are restricted to those where the first die is a 5. The reduced sample space S' is: S' = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}. The total count of outcomes in S' is n(S') = 6.

Next, identify outcomes within S' that satisfy event A, "the sum is greater than 9". Sums for each outcome in S' are:


(5, 1): sum = 6

(5, 2): sum = 7

(5, 3): sum = 8

(5, 4): sum = 9 (not greater than 9)

(5, 5): sum = 10 (greater than 9)

(5, 6): sum = 11 (greater than 9)

Favorable outcomes are {(5, 5), (5, 6)}, totaling 2.

The probability is calculated as the ratio of favorable outcomes to the total outcomes in the reduced sample space.
\[ P(A|B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in reduced sample space}} = \frac{2}{6} = \frac{1}{3} \]

Step 4: Conclusion: The determined probability is $\frac{1{3}$}.