(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 co...
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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.
Answer Details
(a) If \({}^{18}C_r={}^{18}C_{r+2}\), find \({}^{r}C_5\).
Since \({}^{n}C_a={}^{n}C_b\) with \(a\ne b\) requires \(a+b=n\):
\[r+(r+2)=18\;\Rightarrow\;2r+2=18\;\Rightarrow\;r=8\]\[{}^{r}C_5={}^{8}C_5=\frac{8!}{5!\,3!}=56\](b) Binomial with \(p=0.1,\ q=0.9,\ n=6\).
(i) Exactly 5:
\[{}^{6}C_5(0.1)^{5}(0.9)^{1}=6(0.00001)(0.9)=0.000054\approx 0.0001\](ii) None:
\[(0.9)^{6}=0.531441\approx 0.5314\](iii) At most 2 (0, 1 or 2):
\[P(0)=(0.9)^{6}=0.531441\]\[P(1)={}^{6}C_1(0.1)(0.9)^{5}=6(0.1)(0.59049)=0.354294\]\[P(2)={}^{6}C_2(0.1)^{2}(0.9)^{4}=15(0.01)(0.6561)=0.098415\]\[P(\le 2)=0.531441+0.354294+0.098415=0.98415\approx 0.9842\]Download The App On Google Playstore
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