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JEE Advanced · Mathematics · 32. Probability

Paragraph:

Box \(1\) contains three cards bearing numbers \(1,2,3\); box \(2\) contains five cards bearing numbers \(1,2,3,4,5\); and box \(3\) contains seven cards bearing numbers \(1,2,3,4,5,6,7\). A card is drawn from each of the boxes. Let \(x_{i}\) be the number on the card drawn from the \(i^{t h}\) box, \(i=1,2,3\).


Question:

The probability that \(x_{1}+x_{2}+x_{3}\) is odd, is

  1. A 29105
  2. B 53105
  3. C 57105
  4. D 12
Verified Solution

Answer & Solution

Correct Answer

(B) 53105

Step-by-step Solution

Detailed explanation

Case I: One odd, 2 even
Total number of ways =2×2×3+1×3×3+1×2×4=29
Case II: All 3 odd
Number of ways =2×3×4=24
Favourable ways = 53
Required probability =533×5×7=53105
From JEE Advanced
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