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 th...
Question 1 Report
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.
Answer Details
Total balls \(= 12 + 16 + 8 = 36\). Three are drawn without replacement.
(a) All three red:
\[P = \frac{12}{36}\times\frac{11}{35}\times\frac{10}{34} = \frac{1320}{42840} = \frac{11}{357} \approx 0.031\]
(b) First blue, then two reds (in that order):
\[P = \frac{8}{36}\times\frac{12}{35}\times\frac{11}{34} = \frac{1056}{42840} = \frac{44}{1785} \approx 0.025\]
(c) Two white and one blue (in any order): the favourable arrangements are WWB, WBW, BWW. Using selections,
\[P = \frac{\binom{16}{2}\binom{8}{1}}{\binom{36}{3}} = \frac{120 \times 8}{7140} = \frac{960}{7140} = \frac{16}{119} \approx 0.134\]
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