## Applied Statistics and Probability for Engineers, 6th Edition

(a) P(High conductivity ∩ High strength) = $\frac{74}{100}$ = 0.74 (b) Since P(Low conductivity ∩ Low strength) = $\frac{3}{100}$ = 0.03 P(Low conductivity U Low strength) = P(Low conductivity) + P(Low strength) - P(Low conductivity ∩ Low strength) = $\frac{18}{100}$ + $\frac{11}{100}$ - $\frac{3}{100}$ = $\frac{26}{100}$ = 0.26 (c) No, because P(Low conductivity ∩ Low strength) = 0.03 $\ne$ 0