(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.
(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)
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