Question

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?

& \text{A}.\text{ 9}:\text{4} \\\ & \text{B}.\text{ 4}:\text{9} \\\ & \text{C}.\text{ 5}:\text{9} \\\ & \text{D}.\text{ 9}:\text{5} \\\ \end{aligned}$$

Answer 1

Find the probability of drawing a spade from the pack of 52 cards. Then, find the cards and also the probability of drawing an ace from the pack of 52 cards and also the probability of drawing a card which is both spade and ace, which is our ace of spade card. After finding all these probabilities, the probability of drawing a spade or an ace can be found.

Complete step-by-step solution:
A card is drawn at random from a pack of 52 cards and a gamble bets that it is a spade or an ace.
There are 52 cards in the packet, the probability of drawing any one card from the pack is $\dfrac{1}{52}$
We first need to find that the drawn card should be either a spade or an ace.
Let the probability of drawing a spade is P (A).
Let the probability of drawing an ace is P (B).
A well-shuffled pack of 52 cards contains:
13 spades, 13 hearts, 13 diamonds, and 13 clubs.
Each of these 4 categories contains 1 ace, 4 kings, 4 queens, and 4 Js, and other number cards.
Which would make 4 aces in total.
So, P (A) probability of drawing a spade card is $P(A)=\dfrac{13}{52}$
Now, P (B) probability of drawing an ace from 52 cards is $P(B)=\dfrac{4}{52}$
The probability of finding a card which is both spade and ace, that is drawing the ace of spade, which is only one card.
Therefore, $P(A\text{ and B)=}\dfrac{1}{52}$
But we need to find the probability of A or B.
Since,

& P(A\text{ or B)=P(A)+P(B)-P(A and B)} \\\ & \Rightarrow P(A\text{ or B)}=\dfrac{13}{52}+\dfrac{4}{52}-\dfrac{1}{52} \\\ & \Rightarrow P(A\text{ or B)}=\dfrac{13}{52}+\dfrac{3}{52}=\dfrac{16}{52}=\dfrac{4}{13} \\\ & \Rightarrow P(A\text{ or B)}=\dfrac{4}{13} \\\ \end{aligned}$$ We know the total probability is always 1. Since the probability of winning the bet $$\Rightarrow \dfrac{4}{13}$$ Therefore, the probability of losing the bet $$\Rightarrow 1-\dfrac{4}{13}=\dfrac{9}{13}$$ The question asks about the odds against the winning bet, which is the chance of losing the bet over winning it. Simply the ratio of the probability of losing the bet to the probability of winning the bet gives us the required results. **Odds against winning the bet $$\Rightarrow \dfrac{9}{13}:\dfrac{4}{13}\Rightarrow 9:4$$ Therefore, option A is correct.** **Note:** The question has asked to find the odds against winning the bet and not the probability of losing the bet, both of these probabilities are far different. For finding the probability of losing the bet do not go for a long method like P (neither A nor B). $$\text{P }\left( \text{neither A nor B} \right)=\left[ 1-\text{P(A)} \right]+\left[ 1-\text{P(B)} \right]-\left[ 1-\text{P(A and B} \right]$$ Simply do: $$\text{P }\left( \text{neither A nor B} \right)=1-\text{P(A or B)}$$