The well 3

When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% had impurity B. (Obviously, some had both impurities. ) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well:

p0 = no impurities
p1 = exactly one impurity
p2 = both impurities

Correct answer:

p₀ =  20 %
p₁ =  70 %
p₂ =  10 %

Step-by-step explanation:

$$\begin{gathered} P_1=20\%=\frac{20}{100}=\frac{1}{5}=0.2 \\ {P}_2=40\%=\frac{40}{100}=\frac{2}{5}=0.4 \\ {P}_3=50\%=\frac{50}{100}=\frac{1}{2}=0.5 \\ {p}_0=100\cdot P_1=100\cdot 0.2=20\% \end{gathered}$$

$$\begin{gathered} P_4=1-P_1=1-0.2=\frac{4}{5}=0.8 \\ {P}_4=P(A)+P(B)-P(A\cap B) \\ {P}_{12}=P_2+P_3-P_4=0.4+0.5-0.8=\frac{1}{10}=0.1 \\ {p}_1=100\cdot (P_2+P_3-2\cdot P_{12})=100\cdot (0.4+0.5-2\cdot 0.1)=70\% \end{gathered}$$

$$\begin{gathered} p_2=100\cdot P_{12}=100\cdot 0.1=10=10\% \\ \text{ Verifying Solution: } \\ s=p_0+p_1+p_2=20+70+10=100\% \end{gathered}$$

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