Question 18.6: In the draw of a single card from a full deck of playing car...

Equation (18.34) gives P (ace or spade) = P_\text{ace} + P_\text{spade} − P_\text{ace of spades} . Now,

P(\text{A or B}) = P_\text{A} + P_\text{B} −P_{\text{AB}}          (18.34)

P_\text{ace} = \frac{4}{52} = \frac{1}{13},      P_\text{spade} = \frac{13}{52} = \frac{1}{4},     P_\text{ace of spades} = \frac{1}{52}

so

P(\text{ace or spade}) = \frac{4}{52} + \frac{13}{52} – \frac{1}{52} = \frac{16}{52} = 0.308 = 30.8\%