Probability

In a batch of 500 products, there are 10 defective items. In a statistical inspection, 40 products are selected at random from the batch without replacement. The batch is considered satisfactory if there is at most one defective product among the 40 inspected products. Calculate the probability that the batch will be considered satisfactory.

Final Answer:

p =  0.955

Step-by-step explanation:

$$\begin{gathered} p_1=\frac{10}{500}=\frac{1}{50}=0.02 \\ {p}_2=\frac{1}{40}=0.025 \\ {C}_0(40)=\left(\genfrac{}{}{0ex}{}{40}{0}\right)=\frac{40!}{0!(40-0)!}=\frac{1}{1}=1 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{40}{0}\right)\cdot p_1^0\cdot (1-p_1{)}^{40}=1\cdot {\frac{1}{50}}^0\cdot (1-\frac{1}{50}{)}^{40}\doteq 0.4457 \\ {C}_1(40)=\left(\genfrac{}{}{0ex}{}{40}{1}\right)=\frac{40!}{1!(40-1)!}=\frac{40}{1}=40 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{40}{1}\right)\cdot p_1^1\cdot (1-p_1{)}^{39}=40\cdot {\frac{1}{50}}^1\cdot (1-\frac{1}{50}{)}^{39}\doteq 0.3638 \\ p=1-(p_1+p_2)=1-(\frac{1}{50}+\frac{1}{40})=\frac{191}{200}=0.955 \end{gathered}$$

Try another example