<p>An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, …, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is:</p>
<p>\(\dfrac{13}{36}\)</p>
<p>\(\dfrac{15}{72}\)</p>
<p>\(\dfrac{19}{72}\)</p>
<p>\(\dfrac{19}{36}\)</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by conditioning on the coin outcome: P(7 or 8) = P(H)·P(7 or 8|dice) + P(T)·P(7 or 8|cards). Calculate each conditional probability separately, then sum weighted by their respective branch probabilities.
<p><strong>Step 1: Identify the two branches</strong></p><p>The experiment splits into two cases:</p><ul><li>Coin shows Head (probability 1/2): Roll two dice</li><li>Coin shows Tail (probability 1/2): Pick a card from 1-9</li></ul><p><strong>Step 2: Find P(7 or 8 | Dice)</strong></p><p>Ways to get sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways</p><p>Ways to get sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways</p><p>Total favorable outcomes: 6 + 5 = 11 out of 36</p><p>P(7 or 8 | Dice) = 11/36</p><p><strong>Step 3: Find P(7 or 8 | Cards)</strong></p><p>Cards numbered 1-9: only cards 7 and 8 satisfy the condition</p><p>P(7 or 8 | Cards) = 2/9</p><p><strong>Step 4: Apply law of total probability</strong></p><p>P(7 or 8) = P(H) × P(7 or 8|Dice) + P(T) × P(7 or 8|Cards)</p><p>P(7 or 8) = (1/2) × (11/36) + (1/2) × (2/9)</p><p>P(7 or 8) = 11/72 + 2/18</p><p>P(7 or 8) = 11/72 + 8/72 = 19/72</p><p>∴ Answer: C</p>
Correct Answer: C