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Question

(A) Let $A$ be the event that a spade is drawn and $B$ be the event that an ace is drawn.Total number of cards in a pack $= 52$.Number of spades $= 13

Vedclass · Accepted Answer

$9: 4$

Vedclass · Answer

(A) Let $A$ be the event that a spade is drawn and $B$ be the event that an ace is drawn.Total number of cards in a pack $= 52$.Number of spades $= 13$.Number of aces $= 4$.Number of cards that are both a spade and an ace (the ace of spades) $= 1$.Probability of winning the bet $= P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.Probability of losing the bet $= 1 - P(A \cup B) = 1 - \frac{4}{13} = \frac{9}{13}$.Odds against winning the bet are defined as the ratio of the probability of losing to the probability of winning.Odds against $= \frac{9}{13} : \frac{4}{13} = 9 : 4$.

$A$ card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning the bet?

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