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

(a) P(High Scratch Resistance ∩ High Shock Resistance) = $\frac{70}{100}$ = 0.7 (b) P(High Scratch Resistance U High Shock Resistance) = P(High Scratch Resistance) + P(High Shock Resistance) - P(High Scratch Resistance ∩ High Shock Resistance) = $\frac{79}{100}$ + $\frac{86}{100}$ - $\frac{70}{100}$ = $\frac{95}{100}$ = 0.95 (c) No, because P(High Scratch Resistance ∩ High Shock Resistance) = $\frac{70}{100}$ = 0.7 $\ne$ 0