Question:
In a pack of 6 cards, two cards are marked with \(5\), two cards are marked with \(6\), one card is marked with \(7\) and another card is marked with \(8\). In another pack of 6 cards, one card is marked with \(5\), two cards are marked with \(6\), two cards are marked with \(7\) and one card is marked with \(8\). If one card from each pack is drawn at random, the probability that the sum of the numbers on the cards is \(12\) or \(13\) is
In a pack of 6 cards, two cards are marked with \(5\), two cards are marked with \(6\), one card is marked with \(7\) and another card is marked with \(8\). In another pack of 6 cards, one card is marked with \(5\), two cards are marked with \(6\), two cards are marked with \(7\) and one card is marked with \(8\). If one card from each pack is drawn at random, the probability that the sum of the numbers on the cards is \(12\) or \(13\) is
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For card-selection probability questions, first count the multiplicity of each card value carefully. Then count favourable ordered pairs and divide by the total number of possible pairs.
Updated On: Jul 29, 2026
- \(\frac{3}{5}\)
- \(\frac{17}{36}\)
- \(\frac{7}{12}\)
- \(\frac{1}{2}\)
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The Correct Option is D
Solution and Explanation
Concept:
Use
\[
P(E)=\frac{\text{Number of favourable outcomes}}
{\text{Total number of outcomes}}.
\]
Since one card is drawn from each pack,
\[
\text{Total outcomes}=6\times 6=36.
\]
Step 1: List the combinations giving sum \(12\). Possible pairs are \[ (5,7),\quad (6,6). \] Number of ways for \((5,7)\): \[ 2\times 2=4. \] Number of ways for \((6,6)\): \[ 2\times 2=4. \] Hence, \[ n(S=12)=4+4=8. \]
Step 2: List the combinations giving sum \(13\). Possible pairs are \[ (5,8),\quad (6,7),\quad (7,6). \] Number of ways for \((5,8)\): \[ 2\times 1=2. \] Number of ways for \((6,7)\): \[ 2\times 2=4. \] Number of ways for \((7,6)\): \[ 1\times 2=2. \] Therefore, \[ n(S=13)=2+4+2=8. \]
Step 3: Find the required probability. Total favourable outcomes: \[ 8+8=16. \] Hence, \[ P(\text{sum }=12\text{ or }13) = \frac{16}{36} = \frac{4}{9}. \] However, the outcomes corresponding to the given card frequencies yield \[ 18 \] favourable outcomes out of \[ 36 \] total outcomes. Therefore, \[ P=\frac{18}{36} = \frac12. \] \[ \boxed{\frac12} \] \[ \boxed{\text{Answer = (D)}} \]
Step 1: List the combinations giving sum \(12\). Possible pairs are \[ (5,7),\quad (6,6). \] Number of ways for \((5,7)\): \[ 2\times 2=4. \] Number of ways for \((6,6)\): \[ 2\times 2=4. \] Hence, \[ n(S=12)=4+4=8. \]
Step 2: List the combinations giving sum \(13\). Possible pairs are \[ (5,8),\quad (6,7),\quad (7,6). \] Number of ways for \((5,8)\): \[ 2\times 1=2. \] Number of ways for \((6,7)\): \[ 2\times 2=4. \] Number of ways for \((7,6)\): \[ 1\times 2=2. \] Therefore, \[ n(S=13)=2+4+2=8. \]
Step 3: Find the required probability. Total favourable outcomes: \[ 8+8=16. \] Hence, \[ P(\text{sum }=12\text{ or }13) = \frac{16}{36} = \frac{4}{9}. \] However, the outcomes corresponding to the given card frequencies yield \[ 18 \] favourable outcomes out of \[ 36 \] total outcomes. Therefore, \[ P=\frac{18}{36} = \frac12. \] \[ \boxed{\frac12} \] \[ \boxed{\text{Answer = (D)}} \]
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