(a) 9 : 4
Let event A : a spade is drawn and event B : an ace is drawn.
Probability of winning the bet = P (A or B)
P (A or B) = P(A) + P(B) – P (A \(\cap\) B)
Note. Here A and B are not mutually exclusive events, hence the common part has to be taken into consideration.
= \(\frac{13}{52}\)+ \(\frac{4}{52}\) - \(\frac{1}{52}\) (There is one ace of spades)
= \(\frac{16}{52}\) = \(\frac{4}{13}\)
\(\therefore\) Probability of losing the bet = \(1- \frac{4}{13}\)= \(\frac{9}{13}\)
\(\therefore\) Odds against winning the bet = \(\frac{9}{13}\) \(\colon\) \(\frac{4}{13}\) = \(9\colon4\)