(a) If (^{18}C_{r} = ^{18}C_{r + 2}), find (^{r}C_{5}).
(b) In a community, 10% of the people tested positive to the HIV virus. If 6 persons from the community are selected at random, one after the other with replacement, calculate, correct to four decimal places, the probability that : (i) exactly 5 (ii) none (iii) at most 2; tested positive to the virus.
Explanation
(a) (^{18}C_{r} = ^{18}C_{r + 2})
(frac{18!}{r! (18 – r)!} = frac{18!}{(r + 2)! (18 – r – 2)!})
(r! (18 – r)! = (r + 2)(r + 1)r! (18 – r – 2)!)
((18 – r)! = (r + 2)(r + 1)(18 – r – 2)!)
((18 – r)(18 – r – 1)(18 – r – 2)! = (r + 2)(r + 1)(18 – r – 2)!)
((18 – r)(17 – r ) = r^{2} + 3r + 2)
(306 – 35r + r^{2} = r^{2} + 3r + 2)
(306 -2 = r^{2} – r^{2} + 3r + 35r)
(304 = 38r implies r = 8)
(therefore ^{8}C_{5} = frac{8!}{5! (8 – 5)!})
(frac{8 times 7 times 6}{3 times 2} = 56)
(b) p(H) = 10% = 0.1 = a; p(H’) = 0.9 = b
6 persons tested. We use the binomial probability distribution.
((a + b)^{6} = a^{6} + 6a^{5}b + 15a^{4}b^{2} +20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6})
(i) p(exactly 5 tested positive to the virus) = (6a^{5}b)
= (6 times (0.1)^{5} times (0.9) = 0.000054 approxeq 0.0001) (to 4 d.p)
(ii) p(none tested positive) = (b^{6})
= ((0.9)^{6} = 0.531441 approxeq 0.5314) (to 4 d.p)
(iii) p(at most 2) = p(none) + p(one) + p(two)
= (b^{6} + 6ab^{5} + 15a^{2}b^{4})
= ((0.9)^{6} + 6(0.1)(0.9)^{5} + 15(0.1^{2})(0.9^{4}))
= (0.531441 + 0.354294 + 0.098415 = 0.98415)
(approxeq 0.9842) (to 4 d.p)