Certainly! Let's go through the problem step-by-step.
### (a) Probability of Drawing an Ace or a Jack
1. Total Number of Cards: A standard deck has 52 cards.
2. Number of Aces: There are 4 aces in a deck (one for each suit: hearts, diamonds, clubs, spades).
3. Number of Jacks: There are 4 jacks in a deck (one for each suit: hearts, diamonds, clubs, spades).
4. Total Aces or Jacks: The total number of aces or jacks is [tex]\(4 + 4 = 8\)[/tex].
The probability [tex]\( P \)[/tex] of drawing an ace or a jack from the deck is given by the ratio of the number of favorable outcomes to the total number of possible outcomes:
[tex]\[ P(\text{Ace or Jack}) = \frac{\text{Number of Aces or Jacks}}{\text{Total Number of Cards}} = \frac{8}{52} \][/tex]
Simplify this fraction:
[tex]\[ \frac{8}{52} = \frac{4}{26} = \frac{2}{13} \][/tex]
So, the probability that the card is an ace or a jack is:
[tex]\[ 0.15384615384615385 \][/tex]
### (b) Odds in Favor of Drawing an Ace or a Jack
The odds in favor of an event are given by the ratio of the number of favorable outcomes to the number of unfavorable outcomes.
1. Favorable Outcomes: Since there are 8 favorable outcomes (ace or jack), we already know this.
2. Unfavorable Outcomes: The number of unfavorable outcomes is the total number of cards minus the number of favorable outcomes:
[tex]\[ 52 - 8 = 44 \][/tex]
So, the odds in favor of drawing an ace or a jack are:
[tex]\[ \text{Odds in Favor} = \frac{\text{Number of Favorable Outcomes}}{\text{Number of Unfavorable Outcomes}} = \frac{8}{44} \][/tex]
Simplify this fraction:
[tex]\[ \frac{8}{44} = \frac{4}{22} = \frac{2}{11} \][/tex]
So, the odds in favor of the event of drawing an ace or a jack are:
[tex]\[ 0.18181818181818182 \][/tex]
### Summary
(a) The probability that the card is an ace or a jack is:
[tex]\[ \frac{2}{13} \][/tex]
(b) The odds, in simplified form, in favor of the event of the card being an ace or a jack are:
[tex]\[ \frac{2}{11} \][/tex]
