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: Understanding the Concept:
This is a problem of conditional probability. We are given some information (the outcome of the black die) which reduces our sample space, and we need to find the probability of another event within this new, smaller sample space.
Step 2: Key Formula or Approach:
Let A be the event "the sum is greater than 9".
Let B be the event "the black die resulted in a 5".
We need to find P(A|B), the probability of A given B.
The formula is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Alternatively, and more simply, we can directly consider the reduced sample space.
Step 3: Detailed Explanation:
Let's use the reduced sample space method. The event B, "the black die resulted in a 5", has occurred. This means the possible outcomes are now only 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 number of outcomes in this reduced sample space is n(S') = 6.
Now, within this sample space, we want to find the outcomes that satisfy the event A, "the sum is greater than 9". Let's check the sum for each outcome in S':


(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)
The favorable outcomes are {(5, 5), (5, 6)}.
The number of favorable outcomes is 2.
The probability is 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: Final Answer:
The required probability is $\frac{1{3}$}.