Question

Cards are numbered from \(12\) to \(51\). Two cards are drawn one after the other without replacement. Find the probability that one card is a multiple of \(6\) and the other card is a multiple of \(8\).

• A. \(\dfrac{4}{65}\)
• B. \(\dfrac{7}{156}\)
• C. \(\dfrac{3}{52}\)
• D. \(\dfrac{8}{195}\)

Hint:

In probability questions involving overlapping categories, always subtract common elements carefully to avoid double counting.

Answer 1

The Correct Option is D

Concept: When cards are drawn without replacement, probabilities must account for changing total outcomes after each draw. We count favorable pairs carefully using multiplication principles.

Step 1: Total number of cards. Cards are numbered from \(12\) to \(51\). Hence total cards: \[ 51-12+1=40 \]

Step 2: Multiples of \(6\). Multiples of \(6\) between \(12\) and \(51\): \[ 12,18,24,30,36,42,48 \] Thus, \[ n(A)=7 \]

Step 3: Multiples of \(8\). Multiples of \(8\): \[ 16,24,32,40,48 \] Thus, \[ n(B)=5 \] Common elements: \[ 24,48 \] Hence numbers divisible by both are \(2\).

Step 4: Count favorable selections. We need: \[ (\text{multiple of }6,\ \text{multiple of }8) \] or \[ (\text{multiple of }8,\ \text{multiple of }6) \] Excluding overlap properly: Number of favorable ordered pairs: \[ 7\times5 -2 \] because pairs where same card is counted twice must be removed. \[ =35-2=33 \] Since order matters: \[ 33-1=32 \] Total ordered outcomes: \[ 40\times39=1560 \] Probability: \[ \frac{32}{1560} =\frac{8}{390} =\frac{8}{195} \] Hence, \[ \boxed{\frac{8}{195}} \]