(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.

Thursday, 17 April 2025

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(a) If $^{18} C_{r} =^{18} C_{r + 2}$ , find $^{r} C_{5}$ . (b) In a community, 10% of ...

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(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} + 3 r + 2

306 - 35 r + r^{2} = r^{2} + 3 r + 2

306 - 2 = r^{2} - r^{2} + 3 r + 35 r

304 = 38 r ⟹ r = 8

∴^{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} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6}

(i) p(exactly 5 tested positive to the virus) =

6 a^{5} b

6 \times (0.1)^{5} \times (0.9) = 0.000054 ≊ 0.0001

(to 4 d.p)
(ii) p(none tested positive) =

b^{6}

(0.9)^{6} = 0.531441 ≊ 0.5314

(to 4 d.p)
(iii) p(at most 2) = p(none) + p(one) + p(two)
=

b^{6} + 6 a b^{5} + 15 a^{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

≊ 0.9842

(to 4 d.p)

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