Probability - products

Based on previous inspection, it is known that 3% of defective products occur during the manufacturing of a certain product.
a) Calculate the probability that among 100 randomly selected products there are exactly 2 defective ones, assuming each product is returned to the original set after inspection.
b) What is the probability that among 100 randomly selected products there will be at most 2 defective ones?

Final Answer:

p₁ =  0.2252
b =  0.4198

Step-by-step explanation:

$$\begin{gathered} C_2(100)=\left(\genfrac{}{}{0ex}{}{100}{2}\right)=\frac{100!}{2!(100-2)!}=\frac{100\cdot 99}{2\cdot 1}=4950 \\ n=100 \\ q=3\%=\frac{3}{100}=0.03 \\ {p}_1=p_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot q^2\cdot (1-q{)}^{n-2}=4950\cdot 0.03^2\cdot (1-0.03{)}^{100-2}=0.2252 \end{gathered}$$

$$\begin{gathered} C_0(100)=\left(\genfrac{}{}{0ex}{}{100}{0}\right)=\frac{100!}{0!(100-0)!}=\frac{1}{1}=1 \\ {C}_1(100)=\left(\genfrac{}{}{0ex}{}{100}{1}\right)=\frac{100!}{1!(100-1)!}=\frac{100}{1}=100 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot q^0\cdot (1-q{)}^{n-0}=1\cdot 0.03^0\cdot (1-0.03{)}^{100-0}\doteq 0.0476 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot q^1\cdot (1-q{)}^{n-1}=100\cdot 0.03^1\cdot (1-0.03{)}^{100-1}\doteq 0.1471 \\ b=p_0+p_1+p_2=0.0476+0.1471+0.2252=0.4198 \end{gathered}$$

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