Question

The probability for a randomly selected number out of 1, 2, 3, 4, ..., 25 to be a composite number is :

• A. \( \frac{15}{25} \)
• B. \( \frac{10}{25} \)
• C. \( \frac{11}{25} \)
• D. \( \frac{9}{25} \)

Hint:

Always subtract both prime count and '1' from the total to get the composite count.

Answer 1

The Correct Option is A

To solve this problem, we need to determine the probability that a randomly selected number from the set {1, 2, 3, ..., 25} is a composite number. A composite number is a positive integer that has more than two positive divisors.

Let's first identify all the composite numbers in the set:

• The numbers from 1 to 25 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25.
• The numbers that are not composite (i.e., prime numbers and 1) in this range are: 1, 2, 3, 5, 7, 11, 13, 17, 19, 23 (10 numbers in total).

Thus, the composite numbers are those which are not in the above list.

• Composite numbers in this range are: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25 (15 composite numbers in total).

Since there are 25 numbers in total, the probability of selecting a composite number can be calculated as follows:

\(\text{Probability} = \frac{\text{Number of Composite Numbers}}{\text{Total Number of Numbers}}\) \(\text{Probability} = \frac{15}{25}\)

This simplifies to \(\frac{3}{5}\), but considering the provided options, we will keep the fraction as \(\frac{15}{25}\).

Therefore, the probability that a randomly chosen number from 1 to 25 is a composite number is \(\frac{15}{25}\).

The correct answer is: \(\frac{15}{25}\).