Sick days

In Canada, there are typically 261 working days per year. There is a 4.9% chance of an employee taking a sick day.
What is the probability an employee will use 17 OR MORE sick days in a year?

Final Answer:

p =  0.1442

Step-by-step explanation:

$$\begin{gathered} q=4.9\%=\frac{4.9}{100}=0.049 \\ n=261 \\ {C}_0(261)=\left(\genfrac{}{}{0ex}{}{261}{0}\right)=\frac{261!}{0!(261-0)!}=\frac{1}{1}=1 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{261}{0}\right)\cdot q^0\cdot (1-q{)}^{n-0}=1\cdot 0.049^0\cdot (1-0.049{)}^{261-0}\doteq 2.0189\cdot 10^{-6} \\ {C}_1(261)=\left(\genfrac{}{}{0ex}{}{261}{1}\right)=\frac{261!}{1!(261-1)!}=\frac{261}{1}=261 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{261}{1}\right)\cdot q^1\cdot (1-q{)}^{n-1}=261\cdot 0.049^1\cdot (1-0.049{)}^{261-1}\doteq 2.715\cdot 10^{-5} \\ {C}_2(261)=\left(\genfrac{}{}{0ex}{}{261}{2}\right)=\frac{261!}{2!(261-2)!}=\frac{261\cdot 260}{2\cdot 1}=33930 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{261}{2}\right)\cdot q^2\cdot (1-q{)}^{n-2}=33930\cdot 0.049^2\cdot (1-0.049{)}^{261-2}\doteq 0.0002 \\ {C}_3(261)=\left(\genfrac{}{}{0ex}{}{261}{3}\right)=\frac{261!}{3!(261-3)!}=\frac{261\cdot 260\cdot 259}{3\cdot 2\cdot 1}=2929290 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{261}{3}\right)\cdot q^3\cdot (1-q{)}^{n-3}=2929290\cdot 0.049^3\cdot (1-0.049{)}^{261-3}\doteq 0.0008 \\ {C}_4(261)=\left(\genfrac{}{}{0ex}{}{261}{4}\right)=\frac{261!}{4!(261-4)!}=\frac{261\cdot 260\cdot 259\cdot 258}{4\cdot 3\cdot 2\cdot 1}=188939205 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{261}{4}\right)\cdot q^4\cdot (1-q{)}^{n-4}=188939205\cdot 0.049^4\cdot (1-0.049{)}^{261-4}\doteq 0.0027 \\ {C}_5(261)=\left(\genfrac{}{}{0ex}{}{261}{5}\right)=\frac{261!}{5!(261-5)!}=\frac{261\cdot 260\cdot 259\cdot 258\cdot 257}{5\cdot 4\cdot 3\cdot 2\cdot 1}=9711475137 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{261}{5}\right)\cdot q^5\cdot (1-q{)}^{n-5}=9711475137\cdot 0.049^5\cdot (1-0.049{)}^{261-5}\doteq 0.0071 \\ {C}_6(261)=\left(\genfrac{}{}{0ex}{}{261}{6}\right)=\frac{261!}{6!(261-6)!}=\frac{261\cdot 260\cdot 259\cdot 258\cdot 257\cdot 256}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=414356272512 \\ {p}_6=\left(\genfrac{}{}{0ex}{}{261}{6}\right)\cdot q^6\cdot (1-q{)}^{n-6}=414356272512\cdot 0.049^6\cdot (1-0.049{)}^{261-6}\doteq 0.0157 \\ {C}_7(261)=\left(\genfrac{}{}{0ex}{}{261}{7}\right)=\frac{261!}{7!(261-7)!}=\frac{261\cdot 260\cdot 259\cdot 258\cdot 257\cdot 256\cdot 255}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=15094407070080 \\ {p}_7=\left(\genfrac{}{}{0ex}{}{261}{7}\right)\cdot q^7\cdot (1-q{)}^{n-7}=15094407070080\cdot 0.049^7\cdot (1-0.049{)}^{261-7}\doteq 0.0294 \\ {C}_8(261)=\left(\genfrac{}{}{0ex}{}{261}{8}\right)=\frac{261!}{8!(261-8)!}=\frac{261\cdot 260\cdot 259\cdot 258\cdot 257\cdot 256\cdot 255\cdot 254}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=479247424475040 \\ {p}_8=\left(\genfrac{}{}{0ex}{}{261}{8}\right)\cdot q^8\cdot (1-q{)}^{n-8}=479247424475040\cdot 0.049^8\cdot (1-0.049{)}^{261-8}\doteq 0.0481 \\ {C}_9(261)=\left(\genfrac{}{}{0ex}{}{261}{9}\right)=\frac{261!}{9!(261-9)!}=\frac{261\cdot 260\cdot 259\cdot 258\cdot 257\cdot 256\cdot 255\cdot 254\cdot 253}{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=13472177599131680 \\ {p}_9=\left(\genfrac{}{}{0ex}{}{261}{9}\right)\cdot q^9\cdot (1-q{)}^{n-9}=13472177599131680\cdot 0.049^9\cdot (1-0.049{)}^{261-9}\doteq 0.0696 \\ {C}_{10}(261)=\left(\genfrac{}{}{0ex}{}{261}{10}\right)=\frac{261!}{10!(261-10)!}=339498875498118336 \\ {p}_{10}=\left(\genfrac{}{}{0ex}{}{261}{10}\right)\cdot q^{10}\cdot (1-q{)}^{n-10}=339498875498118336\cdot 0.049^{10}\cdot (1-0.049{)}^{261-10}\doteq 0.0904 \\ {C}_{11}(261)=\left(\genfrac{}{}{0ex}{}{261}{11}\right)=\frac{261!}{11!(261-11)!}=7746747068184336576 \\ {p}_{11}=\left(\genfrac{}{}{0ex}{}{261}{11}\right)\cdot q^{11}\cdot (1-q{)}^{n-11}=7746747068184336576\cdot 0.049^{11}\cdot (1-0.049{)}^{261-11}\doteq 0.1063 \\ {C}_{12}(261)=\left(\genfrac{}{}{0ex}{}{261}{12}\right)=\frac{261!}{12!(261-12)!}\approx 1.613\times 10^{20}=161390563920507012000 \\ {p}_{12}=\left(\genfrac{}{}{0ex}{}{261}{12}\right)\cdot q^{12}\cdot (1-q{)}^{n-12}=161390563920507012000\cdot 0.049^{12}\cdot (1-0.049{)}^{261-12}\doteq 0.1141 \\ {C}_{13}(261)=\left(\genfrac{}{}{0ex}{}{261}{13}\right)=\frac{261!}{13!(261-13)!}\approx 3.091\times 10^{21}=3091250032015865076000 \\ {p}_{13}=\left(\genfrac{}{}{0ex}{}{261}{13}\right)\cdot q^{13}\cdot (1-q{)}^{n-13}=3091250032015865076000\cdot 0.049^{13}\cdot (1-0.049{)}^{261-13}\doteq 0.1126 \\ {C}_{14}(261)=\left(\genfrac{}{}{0ex}{}{261}{14}\right)=\frac{261!}{14!(261-14)!}\approx 5.475\times 10^{22}=54759286281423895632000 \\ {p}_{14}=\left(\genfrac{}{}{0ex}{}{261}{14}\right)\cdot q^{14}\cdot (1-q{)}^{n-14}=54759286281423895632000\cdot 0.049^{14}\cdot (1-0.049{)}^{261-14}\doteq 0.1028 \\ {C}_{15}(261)=\left(\genfrac{}{}{0ex}{}{261}{15}\right)=\frac{261!}{15!(261-15)!}\approx 9.017\times 10^{23}=901702914100780148073600 \\ {p}_{15}=\left(\genfrac{}{}{0ex}{}{261}{15}\right)\cdot q^{15}\cdot (1-q{)}^{n-15}=901702914100780148073600\cdot 0.049^{15}\cdot (1-0.049{)}^{261-15}\doteq 0.0872 \\ {C}_{16}(261)=\left(\genfrac{}{}{0ex}{}{261}{16}\right)=\frac{261!}{16!(261-16)!}\approx 1.386\times 10^{25}=13863682304299494776631600 \\ {p}_{16}=\left(\genfrac{}{}{0ex}{}{261}{16}\right)\cdot q^{16}\cdot (1-q{)}^{n-16}=13863682304299494776631600\cdot 0.049^{16}\cdot (1-0.049{)}^{261-16}\doteq 0.0691 \\ s=p_0+p_1+p_2+p_3+p_4+p_5+p_6+p_7+p_8+p_9+p_{10}+p_{11}+p_{12}+p_{13}+p_{14}+p_{15}+p_{16}=2.0189\cdot 10^{-6}+2.715\cdot 10^{-5}+0.0002+0.0008+0.0027+0.0071+0.0157+0.0294+0.0481+0.0696+0.0904+0.1063+0.1141+0.1126+0.1028+0.0872+0.0691\doteq 0.8558 \\ p=1-s=1-0.8558=0.1442 \end{gathered}$$

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