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A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that
(a) All the three balls are white
(b) All the three balls are red
(c) One ball is red and two balls are white
(a) All the three balls are white
(b) All the three balls are red
(c) One ball is red and two balls are white
Solution:
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Verified Answer
(a) $\mathrm{P}$ (all three balls are white $)=\left({ }^5 \mathrm{C}_3{ }^{13} \mathrm{C}_3\right)$
$$
=\frac{\frac{5 !}{3 ! 2 !}}{\frac{13 !}{3 ! 10 !}}=\frac{5.4 .3}{13.12 .11}=\frac{5}{143}
$$
(b) $\mathrm{P}$ (all three balls are red) $=\left({ }^8 \mathrm{C}_3 /{ }^{13} \mathrm{C}_3\right)$
$$
=\frac{\frac{8 !}{3 ! 5 !}}{\frac{13 !}{3 ! 10 !}}=\frac{8.7 .6}{13.12 .11}=\frac{28}{143}
$$
(c) $\mathrm{P}$ (two balls of opposite colour)
$$
=\frac{{ }^8 \mathrm{C}_1 \cdot{ }^5 \mathrm{C}_2}{{ }^{13} \mathrm{C}_3}=\frac{8.10}{\left(\frac{13.12 .11}{3.2 .1}\right)}=\frac{40}{143}
$$
$$
=\frac{\frac{5 !}{3 ! 2 !}}{\frac{13 !}{3 ! 10 !}}=\frac{5.4 .3}{13.12 .11}=\frac{5}{143}
$$
(b) $\mathrm{P}$ (all three balls are red) $=\left({ }^8 \mathrm{C}_3 /{ }^{13} \mathrm{C}_3\right)$
$$
=\frac{\frac{8 !}{3 ! 5 !}}{\frac{13 !}{3 ! 10 !}}=\frac{8.7 .6}{13.12 .11}=\frac{28}{143}
$$
(c) $\mathrm{P}$ (two balls of opposite colour)
$$
=\frac{{ }^8 \mathrm{C}_1 \cdot{ }^5 \mathrm{C}_2}{{ }^{13} \mathrm{C}_3}=\frac{8.10}{\left(\frac{13.12 .11}{3.2 .1}\right)}=\frac{40}{143}
$$
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