Probability 83308

There are 10 parts in the box, and 3 of them are defective. Let's choose 4 components at random. What is the probability that it will be among them
a) 0 defective,
b) just one defective component,
c) just two defective components,
d) exactly 4 defective components?

Correct answer:

p₀ =  0.2401
p₁ =  0.4116
p₂ =  0.2646
p₄ =  0.0081

Step-by-step explanation:

$$\begin{gathered} r=\frac{3}{10}=0.3 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{4}{0}\right)\cdot r^0\cdot (1-r{)}^{4-0}=(\genfrac{}{}{0ex}{}{4}{0})\cdot 0.3^0\cdot (1-0.3{)}^{4-0}=1\cdot 0.3^0\cdot 0.7^4=0.2401 \end{gathered}$$

$p_1=\left(\genfrac{}{}{0ex}{}{4}{1}\right)\cdot r^1\cdot (1-r{)}^{4-1}=(\genfrac{}{}{0ex}{}{4}{1})\cdot 0.3^1\cdot (1-0.3{)}^{4-1}=4\cdot 0.3^1\cdot 0.7^3=0.4116$

$p_2=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot r^2\cdot (1-r{)}^{4-2}=(\genfrac{}{}{0ex}{}{4}{2})\cdot 0.3^2\cdot (1-0.3{)}^{4-2}=6\cdot 0.3^2\cdot 0.7^2=0.2646$

$p_4=\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot r^4\cdot (1-r{)}^{4-4}=(\genfrac{}{}{0ex}{}{4}{4})\cdot 0.3^4\cdot (1-0.3{)}^{4-4}=1\cdot 0.3^4\cdot 0.7^0=0.0081$

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