Component deviation probability

There are 200 components in the production batch, of which 26 have a plus deviation from the nominal value. Calculate the probability that none of the 10 products selected will have a positive variance if we make selections without repetition

Final Answer:

p =  0.2484

Step-by-step explanation:

$$\begin{gathered} C_0(10)=\left(\genfrac{}{}{0ex}{}{10}{0}\right)=\frac{10!}{0!(10-0)!}=\frac{1}{1}=1 \\ n=200 \\ {n}_1=26 \\ {n}_2=10 \\ {p}_1=n_1/n=26/200=\frac{13}{100}=0.13 \\ p=\left(\genfrac{}{}{0ex}{}{n_2}{0}\right)\cdot p_1^0\cdot (1-p_1{)}^{n_2-0}=1\cdot 0.13^0\cdot (1-0.13{)}^{10-0}=0.2484 \end{gathered}$$

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