<p>Four die are thrown simultaneously. The probability that 4 and 3 appear on two of the die given that 5 and 6 have appeared on other two die is</p>
Step-by-Step Solution
Key Concept: This is a conditional probability problem where 5 and 6 are already fixed on two dice, and we need the probability that the remaining two dice show 4 and 3 in any order. Since the outcome of fixed dice doesn't affect the remaining two independent dice, we only focus on those two remaining dice.
<p><strong>Step 1:</strong> Understand the condition - We are given that 5 and 6 have already appeared on two specific dice. We need to find the probability that 4 and 3 appear on the other two dice.</p><p><strong>Step 2:</strong> Since 5 and 6 are fixed on two dice, the remaining two dice are independent and unaffected by this condition. Each remaining die must show either 4 or 3.</p><p><strong>Step 3:</strong> For two remaining dice to show 4 and 3 (in any order):</p><p>- Probability first of the two remaining dice shows 4 and second shows 3: (1/6) × (1/6) = 1/36</p><p>- Probability first shows 3 and second shows 4: (1/6) × (1/6) = 1/36</p><p><strong>Step 4:</strong> Total probability = 1/36 + 1/36 = 2/36 = <strong>1/18</strong></p><p>∴ Answer: A (which is <strong>1/18</strong>)</p>
Correct Answer: A