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A bag contains 12 two rupee coins, 7 one rupee coins and 4 fifty paise coins. If three coins are selected at…
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
$\frac{{ }^4\left({ }^{12} \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_2+{ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_1+{ }^7 \mathrm{C}_2\right)+3\left({ }^{12} \mathrm{C}_1+{ }^7 \mathrm{C}_1\right)}{{ }^{23} \mathrm{C}_3}$ $\frac{4\left({ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_1+{ }^{12} \mathrm{C}_2+{ }^7 \mathrm{C}_2\right)+{ }^3 \mathrm{C}_3}{{ }^{23} \mathrm{C}_3}$ $\frac{4\left({ }^{12} \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_1+{ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_2\right)+3\left({ }^{12} \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_2\right)}{{ }^{23} \mathrm{C}_3}$ $\frac{4\left({ }^{12} \mathrm{C}_3+{ }^7 \mathrm{C}_3\right)+3\left({ }^{12} \mathrm{C}_1+{ }^7 \mathrm{C}_1\right)}{{ }^{23} \mathrm{C}_3}$ Solution
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
$
\begin{aligned}
& \Rightarrow \mathrm{P}=\frac{{ }^4 \mathrm{C}_1{ }^{12} \mathrm{C}_1{ }^7 \mathrm{C}_1{ }^{+12} \mathrm{C}_1{ }^4 \mathrm{C}_1+{ }^7 \mathrm{C}_2{ }^4 \mathrm{C}_1+{ }^4 \mathrm{C}_3}{{ }^{23} \mathrm{C}_3} \\
& \Rightarrow \mathrm{P}=\frac{{ }^4\left({ }^{12} \mathrm{C}_1{ }^7 \mathrm{C}_1{ }^{+12} \mathrm{C}_2{ }^{+7} \mathrm{C}_2\right)+{ }^4 \mathrm{C}_3}{{ }^{23} \mathrm{C}_3}
\end{aligned}
$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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