AP EAMCET · Maths · Probability
A bag contains 12 two rupee coins, 7 one rupee coins and 4 fifty paise coins. If three coins are selected at random, then the probability that the sum of the values of the three coins is not an integral multiple of a rupee is
- A \(\frac{{ }^4\left({ }^{12} \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_2+{ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_1+{ }^7 \mathrm{C}_2\right)+3\left({ }^{12} \mathrm{C}_1+{ }^7 \mathrm{C}_1\right)}{{ }^{23} \mathrm{C}_3}\)
- B \(\frac{4\left({ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_1+{ }^{12} \mathrm{C}_2+{ }^7 \mathrm{C}_2\right)+{ }^3 \mathrm{C}_3}{{ }^{23} \mathrm{C}_3}\)
- C \(\frac{4\left({ }^{12} \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_1+{ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_2\right)+3\left({ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_2\right)}{{ }^{23} \mathrm{C}_3}\)
- D \(\frac{4\left({ }^{12} \mathrm{C}_3+{ }^7 \mathrm{C}_3\right)+3\left({ }^{12} \mathrm{C}_1+{ }^7 \mathrm{C}_1\right)}{{ }^{23} \mathrm{C}_3}\)
Answer & Solution
Correct Answer
(B) \(\frac{4\left({ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_1+{ }^{12} \mathrm{C}_2+{ }^7 \mathrm{C}_2\right)+{ }^3 \mathrm{C}_3}{{ }^{23} \mathrm{C}_3}\)
Step-by-step Solution
Detailed explanation
The following cases are possible: (i) 1 fifty paisa coin, 1 two rupee coin, 1 one rupee coin. (ii) 2 two rupee coin, 1 fifty paisa coin (iii) 2 one rupee coin, 1 fifty paisa coin (iv) 3 fifty paisa coin Then the required probability is…
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