Probability 4665

We have three series of products. We select one product for quality control. Determine the probability of finding a low-quality product if the first batch contains 2/3, the second batch 7/9, and the third batch 3/4 quality products.

Final Answer:

p =  0.2716763

Step-by-step explanation:

$$\begin{gathered} p_1=2/3=\frac{2}{3}\doteq 0.6667 \\ {p}_2=7/9=\frac{7}{9}\doteq 0.7778 \\ {p}_3=3/4=\frac{3}{4}=0.75 \\ {p}_4=3/(1/p_1+1/p_2+1/p_3)=3/(1/0.6667+1/0.7778+1/0.75)=\frac{126}{173}\doteq 0.7283 \\ {p}_5=(p_1+p_2+p_3)/3=(0.6667+0.7778+0.75)/3=\frac{79}{108}\doteq 0.7315 \\ {p}_6=1-p_5=1-0.7315=\frac{29}{108}\doteq 0.2685 \\ p=1-p_4=1-0.7283=0.2716763 \end{gathered}$$

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