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here is a table showing all 52 cards in a standard deck. suppose one ca…

Question

here is a table showing all 52 cards in a standard deck. suppose one card is drawn at random from a standard deck. answer each part. write your answers as fractions in simplest form. (a) what is the probability that the card drawn is a face card? (b) what is the probability that the card drawn is a diamond? (c) what is the probability that the card drawn is a face card or a diamond?

Explanation:

Part (a)

Step1: Identify total and favorable outcomes

A standard deck has 52 cards. Face cards are Jack, Queen, King. Each suit (4 suits: hearts, diamonds, spades, clubs) has 3 face cards, so total face cards = \(4\times3 = 12\).

Step2: Calculate probability

Probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{12}{52}\). Simplify by dividing numerator and denominator by 4: \(\frac{12\div4}{52\div4}=\frac{3}{13}\).

Step1: Identify total and favorable outcomes

Total cards = 52. Diamonds suit has 13 cards (Ace to King).

Step2: Calculate probability

Probability \(P = \frac{\text{Number of diamond cards}}{\text{Total cards}}=\frac{13}{52}\). Simplify by dividing numerator and denominator by 13: \(\frac{13\div13}{52\div13}=\frac{1}{4}\).

Step1: Recall the formula for "or" probability

For two events \(A\) (face card) and \(B\) (diamond), \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

Step2: Find \(P(A)\), \(P(B)\), \(P(A\cap B)\)

  • \(P(A)=\frac{12}{52}\) (from part a), \(P(B)=\frac{13}{52}\) (from part b).
  • \(A\cap B\): face cards in diamonds. Diamonds have 3 face cards (Jack, Queen, King), so \(P(A\cap B)=\frac{3}{52}\).

Step3: Calculate \(P(A\cup B)\)

\(P(A\cup B)=\frac{12}{52}+\frac{13}{52}-\frac{3}{52}=\frac{12 + 13- 3}{52}=\frac{22}{52}\). Simplify by dividing numerator and denominator by 2: \(\frac{11}{26}\).

Answer:

\(\frac{3}{13}\)

Part (b)