A bag contains 4 red, 6 blue and 8 green identical marbles.
(a) If three marbles are drawn at random, without replacement, calculate the probability that :
(i) all will be green ; (ii) all will have the same colour.
(b) If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.
(a) If three marbles are drawn at random, without replacement, calculate the probability that :
(i) all will be green ; (ii) all will have the same colour.
(b) If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.
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Correct Answer: Option n
Explanation:
4 red, 6 blue, 8 green = 18 marbles.
p(red) = \(\frac{4}{18} = \frac{2}{9}\); p(not red) = \(\frac{7}{9}\)
p(blue) = \(\frac{6}{18} = \frac{1}{3}\) ; p(not blue) = \(\frac{2}{3}\)
p(green) = \(\frac{8}{18} = \frac{4}{9}\) ; p(not green) = \(\frac{5}{9}\)
(a) p( 3 green) = \(\frac{8}{18} \times \frac{7}{17} \times \frac{6}{16} = \frac{7}{102}\)
(b) p(all same colour) = p(all red) or p(all blue) or p(all green)
= \(\frac{4}{18} \times \frac{4}{18} \times \frac{4}{18} + \frac{6}{18} \times \frac{6}{18} \times \frac{6}{18} + \frac{8}{18} \times \frac{8}{18} \times \frac{8}{18}\)
= \(\frac{792}{5832}\)
= \(\frac{11}{81}\)
4 red, 6 blue, 8 green = 18 marbles.
p(red) = \(\frac{4}{18} = \frac{2}{9}\); p(not red) = \(\frac{7}{9}\)
p(blue) = \(\frac{6}{18} = \frac{1}{3}\) ; p(not blue) = \(\frac{2}{3}\)
p(green) = \(\frac{8}{18} = \frac{4}{9}\) ; p(not green) = \(\frac{5}{9}\)
(a) p( 3 green) = \(\frac{8}{18} \times \frac{7}{17} \times \frac{6}{16} = \frac{7}{102}\)
(b) p(all same colour) = p(all red) or p(all blue) or p(all green)
= \(\frac{4}{18} \times \frac{4}{18} \times \frac{4}{18} + \frac{6}{18} \times \frac{6}{18} \times \frac{6}{18} + \frac{8}{18} \times \frac{8}{18} \times \frac{8}{18}\)
= \(\frac{792}{5832}\)
= \(\frac{11}{81}\)