Question

eo lesson 5.4 example 2 - finding the probability of overlapping events

1. a bag contains cards numbered 1 through 20. one card is randomly selected. what is the probability that the number on the card is a multiple of 3 or a multiple of 4?

Explanation:

Step1: Identify multiples of 3

Numbers from 1 - 20 that are multiples of 3: 3, 6, 9, 12, 15, 18. So, \( n(3) = 6 \).

Step2: Identify multiples of 4

Numbers from 1 - 20 that are multiples of 4: 4, 8, 12, 16, 20. So, \( n(4) = 5 \).

Step3: Identify common multiples (multiples of LCM(3,4)=12)

Common multiples (multiples of 12) from 1 - 20: 12. So, \( n(3 \cap 4) = 1 \).

Step4: Apply the formula for overlapping events

The formula for \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Here, total number of cards \( N = 20 \).
\( P(3) = \frac{6}{20} \), \( P(4) = \frac{5}{20} \), \( P(3 \cap 4) = \frac{1}{20} \).
So, \( P(3 \cup 4) = \frac{6}{20} + \frac{5}{20} - \frac{1}{20} = \frac{6 + 5 - 1}{20} = \frac{10}{20} = \frac{1}{2} \). Wait, no, 6 + 5 - 1 is 10? Wait 6 + 5 is 11, minus 1 is 10? Wait 6 (multiples of 3) + 5 (multiples of 4) - 1 (common multiple) = 10? Wait let's recount multiples of 3: 3,6,9,12,15,18 (6 numbers). Multiples of 4: 4,8,12,16,20 (5 numbers). Common multiple: 12 (1 number). So total unique numbers: 6 + 5 - 1 = 10. So probability is \( \frac{10}{20} = \frac{1}{2} \)? Wait no, 6 + 5 is 11, minus 1 is 10? Wait 6 + 5 is 11, minus 1 (the overlap) is 10. Yes. So \( \frac{10}{20} = \frac{1}{2} \)? Wait no, 10 divided by 20 is 0.5, but let's check the numbers: multiples of 3 or 4: 3,4,6,8,9,12,15,16,18,20. Wait that's 10 numbers? Wait 3,4,6,8,9,12,15,16,18,20. Let's count: 1.3, 2.4, 3.6, 4.8, 5.9, 6.12, 7.15, 8.16, 9.18, 10.20. Yes, 10 numbers. So probability is \( \frac{10}{20} = \frac{1}{2} \). Wait but let's recalculate the formula: \( P(A \cup B) = \frac{n(A) + n(B) - n(A \cap B)}{N} \). So \( n(A) = 6 \), \( n(B) = 5 \), \( n(A \cap B) = 1 \), \( N = 20 \). So \( \frac{6 + 5 - 1}{20} = \frac{10}{20} = \frac{1}{2} \).

Answer:

\(\frac{1}{2}\)