<p>There are 20 cards. Ten of these cards have the letter "I" printed on them and the other 10 have the letter "T" printed on them. If three cards are picked up at random and kept in the same order, the probability of making word IIT is</p>
Step-by-Step Solution
Key Concept: Calculate probability of drawing cards in a specific sequence (I, then I, then T) without replacement, where order matters and we're selecting from a finite population.
<p><strong>Step 1:</strong> Identify the problem setup. We have 20 cards total: 10 with 'I' and 10 with 'T'. We draw 3 cards in sequence without replacement, keeping them in order.</p><p><strong>Step 2:</strong> For the word 'IIT', we need:</p><ul><li>1st card is 'I': probability = 10/20</li><li>2nd card is 'I': probability = 9/19 (9 'I's remain out of 19 total cards)</li><li>3rd card is 'T': probability = 10/18 (10 'T's remain out of 18 total cards)</li></ul><p><strong>Step 3:</strong> Apply the multiplication rule for sequential events:</p><p>P(IIT) = (10/20) × (9/19) × (10/18)</p><p><strong>Step 4:</strong> Simplify:</p><p>P(IIT) = (10 × 9 × 10)/(20 × 19 × 18) = 900/6840 = 45/342 = 15/114 = 5/38</p><p>∴ Answer: <strong>5/38</strong> (Option D)</p>
Correct Answer: D