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A bag contains 10 similar balls, of which 4 are blue and 6 are red. Three balls are taken out at random from the bag one after the other without replacement. The probabilities that all the three balls drawn are red is
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The correct answer is:
$\frac{1}{6}$
Total no. of balls $=10$
No, of blue balls $=4$
No. of red balls $=6$
No. of ways to draw three red balls $={ }^6 \mathrm{C}_3$
$\therefore$ The required probability is :
$$
\mathrm{P}=\frac{{ }^6 \mathrm{C}_3}{10_{\mathrm{C}_3}}=\frac{\frac{6 \times 5 \times 4}{6}}{\frac{10 \times 9 \times 8}{6}}=\frac{1}{6}
$$
No, of blue balls $=4$
No. of red balls $=6$
No. of ways to draw three red balls $={ }^6 \mathrm{C}_3$
$\therefore$ The required probability is :
$$
\mathrm{P}=\frac{{ }^6 \mathrm{C}_3}{10_{\mathrm{C}_3}}=\frac{\frac{6 \times 5 \times 4}{6}}{\frac{10 \times 9 \times 8}{6}}=\frac{1}{6}
$$
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