A card is drawn from a well shuffled pack of 52 cards. A gambler bets that it is either a heart or an ace. What are odds against his winning this bet?

CONCEPT:

• Let S be a sample space and E be an event such that n(S) = n, n(E) = m and each outcome is equally likely. Then\(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{m}{n} = \frac{{No.\;of\;favourable\;outcomes\;of\;E}}{{Total\;number\;of\;possible\;outcomes}}\)
• If A and B are two events associated with a random experiment, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

CALCULATION:

Let A: Heart is drawn from a pack of cards.

Let B: An ace is drawn from a pack of cards.

As we know that, probability of an event is given by: \(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}}\)

So, probability of winning the bet is given P(A or B) i.e P(A ∪ B)

⇒ P(A) = 13/52, P(B) = 4/52 and P(A ∩ B) = 1/52

As we know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A or B) = P(A ∪ B) = \(\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{4}{13}\)

∴ Probability of losing the bet = \(1 - \frac{4}{13} = \frac{9}{13}\)

∴ Odds against winning the bet is \(\frac{9}{13} : \frac{4}{13} = \frac{9}{4}\)

Hence, option A is the correct answer.