AP EAMCET · Maths · Probability
A basket contains 12 apples in which 3 are rotten. If 3 apples are drawn at random simultaneously from it, then the probability of getting atmost one rotten apple is
- A \(\frac{34}{55}\)
- B \(\frac{48}{55}\)
- C \(\frac{21}{55}\)
- D \(\frac{42}{55}\)
Answer & Solution
Correct Answer
(B) \(\frac{48}{55}\)
Step-by-step Solution
Detailed explanation
Case 1: When No rotten apple is drawn then, No. of ways \(={ }^9 \mathrm{C}_3=84\) Case 2: When 1 rotten apple is drawn then, No. of ways \(={ }^9 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1=\frac{9 \times 8 \times 3}{2}=12 \times 9=108\) Total no of ways of drawing 3 apples…
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