A box contains 40 balls with numbers printed on them from 1 to 40. One ball is drawn at random. Find the probability that it will be a multiple of 3 or 5.
Understand the Problem
The question is asking to calculate the probability of drawing a ball that has a number which is a multiple of either 3 or 5 from a set of 40 balls numbered from 1 to 40. We will identify the multiples of 3 and 5 within the range, ensure to avoid double counting those that are multiples of both, and then determine the probability based on the total number of favorable outcomes over the total number of outcomes.
Answer
The probability is \( \frac{19}{40} \).
Answer for screen readers
The probability that the ball drawn will be a multiple of 3 or 5 is ( \frac{19}{40} ).
Steps to Solve
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Identify the total number of balls There are 40 balls in total, numbered from 1 to 40.
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Count the multiples of 3 The multiples of 3 from 1 to 40 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39 This counts to 13 multiples.
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Count the multiples of 5 The multiples of 5 from 1 to 40 are: 5, 10, 15, 20, 25, 30, 35, 40 This counts to 8 multiples.
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Count the multiples of both 3 and 5 (i.e., multiples of 15) The multiples of 15 from 1 to 40 are: 15, 30 This counts to 2 multiples.
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Use the principle of inclusion-exclusion To find the total number of favorable outcomes, use the formula: $$ \text{Total multiples of 3 or 5} = (\text{Multiples of 3}) + (\text{Multiples of 5}) - (\text{Multiples of both 3 and 5}) $$ Substituting the numbers: $$ \text{Total} = 13 + 8 - 2 = 19 $$
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Calculate the probability The probability of drawing a ball that is a multiple of 3 or 5 is calculated as: $$ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{19}{40} $$
The probability that the ball drawn will be a multiple of 3 or 5 is ( \frac{19}{40} ).
More Information
The principle of inclusion-exclusion helps avoid double counting when finding the total number of favorable outcomes. In this problem, it ensured that numbers counted as multiples of both 3 and 5 were not counted twice.
Tips
- Forgetting to subtract the common multiples can lead to overcounting.
- Miscounting the multiples of either 3 or 5 due to overlooking some numbers.
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