Probability of failures

The probability of failure in specific productions is 0.01. Calculate the likelihood of more than one failure among the 100 selected products if we return them to the file after the check.

Final Answer:

p =  0.2642

Step-by-step explanation:

$$\begin{gathered} C_0(100)=\left(\genfrac{}{}{0ex}{}{100}{0}\right)=\frac{100!}{0!(100-0)!}=\frac{1}{1}=1 \\ {C}_1(100)=\left(\genfrac{}{}{0ex}{}{100}{1}\right)=\frac{100!}{1!(100-1)!}=\frac{100}{1}=100 \\ q=0.01 \\ n=100 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot q^0\cdot (1-q{)}^{n-0}=1\cdot 0.01^0\cdot (1-0.01{)}^{100-0}\doteq 0.366 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot q^1\cdot (1-q{)}^{n-1}=100\cdot 0.01^1\cdot (1-0.01{)}^{100-1}\doteq 0.3697 \\ p=1-(p_0+p_1)=1-(0.366+0.3697)=0.2642 \end{gathered}$$

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