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 problem involves conditional probability. We are given an observation (the black die's result) that narrows the possible outcomes and requires calculating a probability within this restricted set.
Step 2: Method Identification:
Let A be the event "sum exceeds 9".
Let B be the event "black die shows 5".
We aim 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, a direct calculation using the reduced sample space is simpler.
Step 3: Detailed Analysis:
Employing the reduced sample space method:
Event B, "black die shows 5", has occurred. The outcomes are now limited to those where the first die is 5.
Reduced sample space S':
S' = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
The size of this reduced sample space is n(S') = 6.
Now, we identify outcomes within S' that satisfy event A, "sum exceeds 9". Sums for each outcome in S':


(5, 1): sum = 6

(5, 2): sum = 7

(5, 3): sum = 8

(5, 4): sum = 9 (not>9)

(5, 5): sum = 10 (> 9)

(5, 6): sum = 11 (> 9)

Favorable outcomes are {(5, 5), (5, 6)}.
The count of favorable outcomes is 2.

The probability is the ratio of favorable outcomes to the total outcomes in the reduced sample space:
\[ P(A|B) = \frac{\text{Favorable outcomes}}{\text{Total outcomes in reduced sample space}} = \frac{2}{6} = \frac{1}{3} \]
Step 4: Conclusion:
The calculated probability is $\frac{1}{3}$.