A die is thrown two times and the sum of the scores appearing on the die is observed to be a multipl

Q: Why is the denominator 9 and not 36? I thought the total outcomes were 36.

That is a classic trap! When we say 'given the sum is a multiple of 4', we are restricting our sample space. We are no longer looking at all 36 outcomes; we are only looking at the subset where the sum is 4, 8, or 12. This is the essence of conditional probability—the denominator shrinks to the size of the condition.

Q: How do I know I haven't missed any pairs for the sum of 8?

A systematic approach is your best friend. Start with the smallest possible value for the first die that could lead to a sum of 8. Since $1+6=7$, the first die must be at least 2. Then, for each value of the first die (2, 3, 4, 5, 6), check if the required value for the second die is valid (between 1 and 6). This ensures you catch every single pair without guessing.

Q: Why is the answer 1/9? Does the order of the dice matter?

Yes, the order matters! In probability, $(1,3)$ and $(3,1)$ are distinct outcomes. Even though the sum is the same, they represent different physical events. By treating them as distinct, we ensure our probability calculations are accurate and consistent with the fundamental counting principle.

Question

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

Options:

(A) 1/3

(B) 1/4

(C) 1/8

(D) 1/9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast, empty grid representing the universe of possibilities when rolling two six-sided dice. With six faces on each die, the total number of outcomes is $6 	imes 6 = 36$. This represents our total sample space.

However, we are interested in a restricted reality where the sum of the scores is a multiple of 4. This condition defines Event $B$. Consequently, the vast majority of our 36 outcomes vanish, leaving only those where the sum is 4, 8, or 12.

### Defining the Condition

To determine the size of our reduced sample space, $n(B)$, we systematically identify the pairs that satisfy the condition:

For a sum of 4, the pairs are $(1, 3), (2, 2),$ and $(3, 1)$. This yields 3 outcomes.

For a sum of 8, we identify $(2, 6), (3, 5), (4, 4), (5, 3),$ and $(6, 2)$. This yields 5 outcomes.

For a sum of 12, there is only one possibility: $(6, 6)$. This yields 1 outcome.

Summing these, we find the total size of our restricted sample space:
$$n(B) = 3 + 5 + 1 = 9$$

### The Intersection: Finding the Target

Now, we introduce Event $A$, defined as the event where the score 4 appears at least once. We must find the intersection $n(A \cap B)$, which requires us to count how many outcomes within our restricted set of 9 contain the number 4.

Scanning our list:
- From the sum of 4: $(1, 3), (2, 2), (3, 1)$ contain no 4.
- From the sum of 8: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$, we find that $(4, 4)$ contains a 4.
- From the sum of 12: $(6, 6)$ contains no 4.

Thus, the number of favorable outcomes is:
$$n(A \cap B) = 1$$

### The Final Calculation

The formula for conditional probability is given by:
$$P(A|B) = \frac{n(A \cap B)}{n(B)}$$

Substituting our calculated values into the formula, we arrive at the final result:
$$P(A|B) = \frac{1}{9}$$

This result demonstrates that probability is about visualizing constraints and counting with precision. By navigating the grid and isolating the favorable outcomes, we have successfully determined the **final answer: $\frac{1}{9}$**.

Animated Solution for Mathematics - Probability: A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

Select Answer:

Visualized Solution

Visualizing the Sample Space

Defining the Condition (Event $B$ )

Finding Outcomes for Sum $= 4$

Finding Outcomes for Sum $= 8$

Finding Outcomes for Sum $= 12$

Calculating Total Sample Space $n (B)$

Identifying Favorable Outcomes ( $A \cap B$ )

Final Probability Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

Defining the Condition

The Intersection

The Final Calculation

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Animated Solution for Mathematics - Probability: A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

Select Answer:

Visualized Solution

Visualizing the Sample Space

Defining the Condition (Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B)

Finding Outcomes for Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } =4

Finding Outcomes for Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } =8

Finding Outcomes for Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } =12

Calculating Total Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(B)

Identifying Favorable Outcomes ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩B)