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a standard deck of cards has 13 ranks: 2 - 10, jack, queen, king and ac…

Question

a standard deck of cards has 13 ranks: 2 - 10, jack, queen, king and ace. it has 4 suits: diamonds and hearts (red), and spades and clubs (black). enter all your answers as fractions. if you draw a card at random, whats the probability it is an ace? whats the probability its not an ace? whats the probability its an ace or a black card? (hint: use the idea from the union rule to count up the number of aces with the number of black cards, and subtract the ones that are both!) whats the probability its an ace or a king? whats the probability its an ace and a king?

Explanation:

Step1: Determine total number of cards

A standard deck has 52 cards.

Step2: Calculate probability of drawing an Ace

There are 4 Aces in a deck. Probability $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{4}{52}=\frac{1}{13}$.

Step3: Calculate probability of not drawing an Ace

The probability of not - A is $P(\text{not }A)=1 - P(A)=1-\frac{1}{13}=\frac{12}{13}$.

Step4: Calculate probability of Ace or black card

Number of black cards $n(B)=26$, number of Aces $n(A) = 4$, number of black Aces $n(A\cap B)=2$. By the union rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{4}{52}+\frac{26}{52}-\frac{2}{52}=\frac{4 + 26-2}{52}=\frac{28}{52}=\frac{7}{13}$.

Step5: Calculate probability of Ace or King

Number of Aces $n(A)=4$, number of Kings $n(K)=4$, number of cards that are both Ace and King $n(A\cap K)=0$. By the union rule $P(A\cup K)=P(A)+P(K)-P(A\cap K)=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}=\frac{2}{13}$.

Step6: Calculate probability of Ace and King

A card cannot be both an Ace and a King simultaneously, so $P(A\cap K) = 0=\frac{0}{52}$.

Answer:

  1. $\frac{1}{13}$
  2. $\frac{12}{13}$
  3. $\frac{7}{13}$
  4. $\frac{2}{13}$
  5. $\frac{0}{52}$