Q78 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics
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$\frac{34}{55}$
- Correct answerB$\frac{48}{55}$
- C$\frac{21}{55}$
- D$\frac{42}{55}$
Solution
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 $={ }^{12} \mathrm{C}_3$ $\text { Required probability }=\frac{(84+108)}{{ }^{12} \mathrm{C}_3}=\frac{48}{55}$
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 $={ }^{12} \mathrm{C}_3$ $\text { Required probability }=\frac{(84+108)}{{ }^{12} \mathrm{C}_3}=\frac{48}{55}$
Hence, the correct answer is option: 'B'.