Contamination problems in semiconductor manufacturing can result in a functional defect, a minor defect, or no defect in the final product. Suppose that, 20, 50 and 30% of the contamination problems result functional, minor and no defects respectively. Assume that the effects of 10 contamination problems are independent. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that the 10 contamination problems result in two functional defects and five minor defects

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Answer:

The answer is "[tex]8.5\times 10^{-9}[/tex]".

Step-by-step explanation:

Following are the calculation of the multinomial distribution probability:

[tex]\to (\frac{10!}{(2! \times 5! \times 3!)})\times(0.2)^2 \times(0.5)^5 \times (0.3)^3 \\\\ \to (\frac{10!}{(2! \times 5! \times 3!)})\times 0.04 \times 0.03125 \times 0.027\\\\ \to (\frac{10\times9\times8 \times7 \times6 \times5! }{(2! \times 5! \times 3!)})\times 3.37 \times 10^{-5}\\\\ \to (\frac{10\times9\times8 \times7 \times6 }{(2 \times 1 \times 3 \times 2 \times 1 )})\times 3.37 \times 10^{-5}\\\\[/tex]

[tex]\to (\frac{10\times9\times8 \times7 }{(2)})\times 3.37 \times 10^{-5}\\\\ \to (5 \times 9 \times 8 \times 7 )\times 3.37 \times 10^{-5}\\\\ \to 45 \times 56 \times 3.37 \times 10^{-5}\\\\ \to 45 \times 56 \times 3.37 \times 10^{-5}\\\\ \to 2520 \times 3.37 \times 10^{-5}\\\\\to 8492.4 \times 10^{-5}\\\\\to 8.4924\times 10^{-9}\\\\ \to 8.5\times 10^{-9}[/tex]