A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :
(a) three are red;
(b) the first is blue and the other two are red;
(c) two are white and one is blue.
(a) three are red;
(b) the first is blue and the other two are red;
(c) two are white and one is blue.
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Correct Answer: Option n
Explanation:
n(R) = number of red balls = 12;
n(W) = number of white balls = 16;
n(B) = number of blue balls = 8
Total number of balls = 36
(a) P(three are red) = \(\frac{12}{36} \times \frac{11}{35} \times \frac{10}{34}\)
= \(\frac{11}{357}\)
(b) P(first is blue and other two red) = \(\frac{8}{36} \times \frac{12}{35} \times \frac{11}{34}\)
= \(\frac{44}{1785}\)
(c) P(two are white and one is blue) = \(\frac{16}{36} \times \frac{15}{35} \times \frac{8}{34}\)
= \(\frac{16}{357}\)
n(R) = number of red balls = 12;
n(W) = number of white balls = 16;
n(B) = number of blue balls = 8
Total number of balls = 36
(a) P(three are red) = \(\frac{12}{36} \times \frac{11}{35} \times \frac{10}{34}\)
= \(\frac{11}{357}\)
(b) P(first is blue and other two red) = \(\frac{8}{36} \times \frac{12}{35} \times \frac{11}{34}\)
= \(\frac{44}{1785}\)
(c) P(two are white and one is blue) = \(\frac{16}{36} \times \frac{15}{35} \times \frac{8}{34}\)
= \(\frac{16}{357}\)