Manufacturer 80810

The manufacturer indicates that the germination rate of the pepper seeds is 68%. What is the probability that
a) out of ten seeds sown, at least 8 will germinate?
b) will at most 3 sprouts out of ten seeds be sown?

Correct answer:

a =  0.3313
b =  0.0155

Step-by-step explanation:

$$\begin{gathered} k=68\%=\frac{68}{100}=\frac{17}{25}=0.68 \\ n=10 \\ {p}_8=\left(\genfrac{}{}{0ex}{}{n}{8}\right)\cdot k^8\cdot (1-k{)}^{n-8}=(\genfrac{}{}{0ex}{}{10}{8})\cdot 0.68^8\cdot (1-0.68{)}^{10-8}=45\cdot 0.68^8\cdot 0.32^2\doteq 0.2107 \\ {p}_9=\left(\genfrac{}{}{0ex}{}{n}{9}\right)\cdot k^9\cdot (1-k{)}^{n-9}=(\genfrac{}{}{0ex}{}{10}{9})\cdot 0.68^9\cdot (1-0.68{)}^{10-9}=10\cdot 0.68^9\cdot 0.32^1\doteq 0.0995 \\ {p}_{10}=\left(\genfrac{}{}{0ex}{}{n}{10}\right)\cdot k^{10}\cdot (1-k{)}^{n-10}=(\genfrac{}{}{0ex}{}{10}{10})\cdot 0.68^{10}\cdot (1-0.68{)}^{10-10}=1\cdot 0.68^{10}\cdot 0.32^0\doteq 0.0211 \\ a=p_8+p_9+p_{10}=0.2107+0.0995+0.0211=0.3313 \end{gathered}$$

$$\begin{gathered} p_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot k^0\cdot (1-k{)}^{n-0}=(\genfrac{}{}{0ex}{}{10}{0})\cdot 0.68^0\cdot (1-0.68{)}^{10-0}=1\cdot 0.68^0\cdot 0.32^{10}\doteq 1.1259\cdot 10^{-5} \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot k^1\cdot (1-k{)}^{n-1}=(\genfrac{}{}{0ex}{}{10}{1})\cdot 0.68^1\cdot (1-0.68{)}^{10-1}=10\cdot 0.68^1\cdot 0.32^9\doteq 0.0002 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot k^2\cdot (1-k{)}^{n-2}=(\genfrac{}{}{0ex}{}{10}{2})\cdot 0.68^2\cdot (1-0.68{)}^{10-2}=45\cdot 0.68^2\cdot 0.32^8\doteq 0.0023 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot k^3\cdot (1-k{)}^{n-3}=(\genfrac{}{}{0ex}{}{10}{3})\cdot 0.68^3\cdot (1-0.68{)}^{10-3}=120\cdot 0.68^3\cdot 0.32^7\doteq 0.013 \\ b=p_0+p_1+p_2+p_3=1.1259\cdot 10^{-5}+0.0002+0.0023+0.013=0.0155 \end{gathered}$$

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