Probability of Drawing a Ticket Number Multiple of 3 or 5
In this article, we'll explore a classic problem involving probability: determining the likelihood of drawing a ticket numbered between 1 and 120 that is a multiple of either 3 or 5. We'll solve this problem using the principle of inclusion-exclusion and provide a step-by-step explanation.
Introduction to the Problem
The problem involves a bag containing 120 tickets, each numbered from 1 to 120. We need to find the probability that a randomly drawn ticket has a number that is a multiple of 3 or 5. This is a fundamental exercise in probability theory that can be applied to many real-world scenarios, such as statistical surveys or lottery analysis.
Multiples of 3
First, let's count the multiples of 3 within the range from 1 to 120.
The first multiple of 3 is 3, and the last is 120. This sequence forms an arithmetic series with the first term a 3, common difference d 3, and the n-th term given by a_n 3n. We need to find n such that 3n ≤ 120. Solving for n, we get n ≤ 120 / 3 40.Therefore, there are 40 multiples of 3 in the range from 1 to 120.
Multiples of 5
Next, we count the multiples of 5 within the same range.
The first multiple of 5 is 5, and the last is 120. This sequence also forms an arithmetic series with the first term a 5, common difference d 5. We need to find n such that 5n ≤ 120. Solving for n, we get n ≤ 120 / 5 24.Hence, there are 24 multiples of 5 in the range from 1 to 120.
Multiples of Both 3 and 5 (i.e., 15)
To avoid double-counting, we count the multiples of both 3 and 5, which are multiples of 15.
The first multiple of 15 is 15, and the last is 120. This sequence forms an arithmetic series with the first term a 15, common difference d 15. We need to find n such that 15n ≤ 120. Solving for n, we get n ≤ 120 / 15 8.Thus, there are 8 multiples of 15 in the range from 1 to 120.
Applying the Principle of Inclusion-Exclusion
The principle of inclusion-exclusion helps us prevent double-counting the multiples of both 3 and 5.
The formula to find the total number of favorable outcomes for multiples of 3 or 5 is:
Number of multiples of 3 or 5 Number of multiples of 3 Number of multiples of 5 - Number of multiples of 15
Substituting our values:
Number of multiples of 3 or 5 40 24 - 8 56
Calculating the Probability
There are 120 tickets in total, and 56 of them are multiples of 3 or 5.
The probability of drawing a ticket that is a multiple of 3 or 5 is given by:
P Number of favorable outcomes / Total number of outcomes 56 / 120
Simplifying this fraction:
P (56 / 8) / (120 / 8) 7 / 15
Therefore, the probability is 7/15.
Related Examples
Consider a smaller example for better understanding. If there are 20 tickets, where 3, 6, 9, 12, 15, and 18 are multiples of 3 and 5, 10, 15, and 20 are multiples of 5, and the common multiples are 15 (which repeats in both), the probability would be:
Total favorable outcomes 6 (multiples of 3) 4 (multiples of 5) - 2 (common multiples) 8 Total outcomes 20 Probability 8 / 20 0.45Conclusion
In conclusion, the probability of drawing a ticket numbered between 1 and 120 that is a multiple of 3 or 5 is 7/15. Understanding this principle is crucial for solving a wide range of probability problems and has practical applications in various fields such as statistics, analytics, and finance.