A card is drawn from an ordinary pack of 52 cards and a gambler bets t

1.	A card is drawn from an ordinary pack of 52 cards and a gambler bets that it is a spade or an ace. What are the odds against his winning the bet? (a) 9 : 4 (b) 4 : 9 (c) 5 : 9 (d) 9 : 5
Answer» (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 = \(\frac{9}{13}\) \(\colon\) \(\frac{4}{13}\) = \(9\colon4\)

A card is drawn from an ordinary pack of 52 cards and a gambler bets that it is a spade or an ace. What are the odds against his winning the bet? (a) 9 : 4 (b) 4 : 9 (c) 5 : 9 (d) 9 : 5

Answer»

(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 = \(\frac{9}{13}\) \(\colon\) \(\frac{4}{13}\) = \(9\colon4\)

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