Question

from a bag of 20 cards, whose faces are numbered 1 to 20, one card is drawn at random. what is the probability of getting a multiple of 3 or a multiple of 4?

Explanation:

Step1: Identify multiples of 3 and 4

Multiples of 3 between 1 - 20: 3, 6, 9, 12, 15, 18 (count = 6)
Multiples of 4 between 1 - 20: 4, 8, 12, 16, 20 (count = 5)
Common multiple (12) is counted twice, so total unique favorable outcomes: \(6 + 5 - 1 = 10\)

Step2: Calculate probability

Total cards = 20. Probability = \(\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{10}{20} = \frac{1}{2}\)? Wait, no—wait, 6 + 5 -1 = 10? Wait 6 (multiples of 3) + 5 (multiples of 4) -1 (overlap: 12) = 10? Wait 3,6,9,12,15,18 (6 numbers); 4,8,12,16,20 (5 numbers). Overlap is 12, so unique is 6 + 5 -1 = 10. Then probability is \( \frac{10}{20} = \frac{1}{2} \)? Wait, but let's check again. Wait 3,6,9,12,15,18 (6); 4,8,12,16,20 (5). So combined: 3,4,6,8,9,12,15,16,18,20. That's 10 numbers. So 10/20 = 1/2. Wait, but wait the options include 3/5 (12/20), 1/2 (10/20), etc. Wait, did I miscalculate? Wait 6 + 5 = 11, minus 1 (overlap) is 10. So 10/20 = 1/2. So the probability is \( \frac{1}{2} \).

Wait, no—wait 3,6,9,12,15,18 (6); 4,8,12,16,20 (5). So the union is {3,4,6,8,9,12,15,16,18,20} → 10 elements. So 10/20 = 1/2. So the answer is \( \frac{1}{2} \).

Answer:

\(\frac{1}{2}\) (corresponding to the option with \( \frac{1}{2} \))