The problem in the image asks to find the probability that a randomly selected disc from a box of 20 discs, numbered consecutively from 1 to 20, has a number divisible by 3 or 5.

Step 1: List numbers divisible by 3 or 5

We will first list the numbers from 1 to 20 that are divisible by 3 and those divisible by 5:

• Divisible by 3: 3, 6, 9, 12, 15, 18 (6 numbers)
• Divisible by 5: 5, 10, 15, 20 (4 numbers)

Step 2: Consider the overlap

The number 15 is divisible by both 3 and 5, so it has been counted twice in both lists. To avoid double-counting, we need to subtract this overlap.

Step 3: Use the principle of inclusion-exclusion

• Total numbers divisible by 3 or 5 = (Numbers divisible by 3) + (Numbers divisible by 5) − (Numbers divisible by both)
• Total numbers divisible by 3 or 5 = 6 + 4 − 1 = 9

Step 4: Find the probability

The probability is the ratio of favorable outcomes (numbers divisible by 3 or 5) to the total outcomes (total discs): $P(\text{divisible by 3 or 5}) = \frac{9}{20}$

Thus, the probability that the number on the randomly selected disc is divisible by 3 or 5 is $\frac{9}{20}$.

Would you like further details or have any questions? Here are five related questions to expand on this concept:

1. How would the probability change if the range of numbers was from 1 to 30?
2. What is the probability of selecting a number divisible by both 2 and 3?
3. How many numbers between 1 and 100 are divisible by 3 or 5?
4. What if the discs were numbered from 1 to 50—how would that affect the probability?
5. Can you generalize the probability for any given range divisible by two numbers?

Tip: When calculating the probability for multiple conditions (like divisibility by 3 or 5), remember to account for overlapping cases using the inclusion-exclusion principle!