Question

A number is chosen randomly from each of the two following sets 
Set 1: {1,2,3,4,5,6,7 }
Set 2: {3,4,5,6,7,8} 
Find the probability that the sum of the two numbers is 6. 
 

• A. $\frac{3}{16}$
• B. $\frac{1}{14}$
• C. $\frac{1}{7}$
• D. $\frac{1}{6}$

Hint:

Explicitly write down the conditions for the second element to ensure you do not include values that fall outside the boundary of the second set.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space.
Key Formula or Approach:
For two independent selections from finite sets $A$ and $B$:
\[ \text{Total Outcomes} = n(A) \times n(B) \]
\[ P(\text{Sum} = 6) = \frac{\text{Number of pairs } (a, b) \text{ such that } a + b = 6}{\text{Total Outcomes}} \]

Step 2: Detailed Explanation:
Let us determine the size of the sample space:
Set 1, $A = \{1, 2, 3, 4, 5, 6, 7\} \implies n(A) = 7$
Set 2, $B = \{3, 4, 5, 6, 7, 8\} \implies n(B) = 6$
Total possible outcomes = $7 \times 6 = 42$ pairs.
We want to find pairs $(a, b)$ with $a \in A$ and $b \in B$ such that $a + b = 6$.
Let us check the possibilities based on elements in Set 1 ($a$):
- If $a = 1$: we need $b = 5$. Since $5 \in B$, $(1, 5)$ is a valid pair.
- If $a = 2$: we need $b = 4$. Since $4 \in B$, $(2, 4)$ is a valid pair.
- If $a = 3$: we need $b = 3$. Since $3 \in B$, $(3, 3)$ is a valid pair.
- If $a = 4$: we need $b = 2$. However, $2 \notin B$, so this is invalid.
- For any $a \ge 4$, the required value of $b$ to get a sum of 6 would be $\le 2$, which does not exist in Set 2.
Thus, the favorable pairs are only: $\{(1, 5), (2, 4), (3, 3)\}$.
Number of favorable outcomes = 3.
Compute the probability:
\[ P(\text{Sum} = 6) = \frac{3}{42} = \frac{1}{14} \]
Therefore, the probability is $\frac{1}{14}$.

Step 3: Final Answer
The correct option is (B).