Question

there are 4 jacks and 13 clubs in a standard 52 - card deck of playing cards. what is the probability that a card picked at random from a standard deck of playing cards is a club or a jack?
a. \\(\frac{4}{13}\\)
b. \\(\frac{1}{22}\\)
c. \\(\frac{3}{13}\\)
d. \\(\frac{7}{26}\\)

Explanation:

Step1: Recall the formula for probability of union

The formula for the probability of \( A \) or \( B \) (i.e., \( P(A \cup B) \)) is \( P(A) + P(B) - P(A \cap B) \), where \( A \) is the event of picking a club and \( B \) is the event of picking a jack.

Step2: Calculate \( P(A) \), \( P(B) \), and \( P(A \cap B) \)

• \( P(A) \): There are 13 clubs in 52 cards, so \( P(A)=\frac{13}{52} \).
• \( P(B) \): There are 4 jacks in 52 cards, so \( P(B)=\frac{4}{52} \).
• \( P(A \cap B) \): The jack of clubs is the overlap, so there is 1 such card. Thus, \( P(A \cap B)=\frac{1}{52} \).

Step3: Apply the formula

Substitute the values into the formula: \( P(A \cup B)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{13 + 4 - 1}{52}=\frac{16}{52}=\frac{4}{13} \)? Wait, no, wait: 13 + 4 is 17, minus 1 is 16? Wait, no, 13 clubs, 4 jacks, but one jack is a club (jack of clubs), so the number of cards that are club or jack is \( 13 + 4 - 1 = 16 \)? Wait, no, 13 clubs: 13 cards. 4 jacks: but one jack is already in clubs, so the unique cards are 13 (clubs) + 3 (jacks that are not clubs) = 16. Then probability is \( \frac{16}{52}=\frac{4}{13} \)? Wait, but let's check the options. Option A is \( \frac{4}{13} \), but wait, let's recalculate:

Wait, 13 clubs, 4 jacks, but the jack of clubs is counted in both. So the number of favorable outcomes is \( 13 + 4 - 1 = 16 \). Total outcomes 52. So \( \frac{16}{52}=\frac{4}{13} \). But let's check the options again. Wait, maybe I made a mistake. Wait, 13 clubs: 13, 4 jacks: but one is a club, so 13 + 3 = 16. 16/52 simplifies to 4/13 (divide numerator and denominator by 4: 16÷4=4, 52÷4=13). So the answer should be A. \( \frac{4}{13} \).

Answer:

A. \( \frac{4}{13} \)