A number is selected at random from the integers 30 to 48 inclusive. We want to find the probability that the number is (a) divisible by 3, (b) prime, and (c) prime or divisible by 3. The answers should be rounded to four significant figures.

Probability and StatisticsProbabilityPrime NumbersDivisibility
2025/6/3

1. Problem Description

A number is selected at random from the integers 30 to 48 inclusive. We want to find the probability that the number is (a) divisible by 3, (b) prime, and (c) prime or divisible by

3. The answers should be rounded to four significant figures.

2. Solution Steps

First, we determine the total number of possible outcomes. The integers range from 30 to 48, so the total number of integers is 4830+1=1948 - 30 + 1 = 19.
(a) Divisible by 3
We need to find the number of integers between 30 and 48 that are divisible by

3. The first such integer is 30 and the last is

4

8. The integers divisible by 3 are $30, 33, 36, 39, 42, 45, 48$. There are 7 such integers.

Therefore, the probability that the number selected is divisible by 3 is 719\frac{7}{19}.
7190.368421052630.3684\frac{7}{19} \approx 0.36842105263 \approx 0.3684 (to four significant figures).
(b) Prime
We need to find the number of prime numbers between 30 and
4

8. A prime number is a number greater than 1 that has only two factors: 1 and itself. The prime numbers in this range are $31, 37, 41, 43, 47$. There are 5 prime numbers.

Therefore, the probability that the number selected is prime is 519\frac{5}{19}.
5190.263157894730.2632\frac{5}{19} \approx 0.26315789473 \approx 0.2632 (to four significant figures).
(c) Prime or divisible by 3
We need to find the number of integers between 30 and 48 that are either prime or divisible by

3. The integers divisible by 3 are $30, 33, 36, 39, 42, 45, 48$.

The prime numbers are 31,37,41,43,4731, 37, 41, 43, 47.
The set of integers that are prime or divisible by 3 is {30,31,33,36,37,39,41,42,43,45,47,48}\{30, 31, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48\}. There are 12 such integers.
Therefore, the probability that the number selected is prime or divisible by 3 is 1219\frac{12}{19}.
12190.631578947360.6316\frac{12}{19} \approx 0.63157894736 \approx 0.6316 (to four significant figures).

3. Final Answer

(a) 0.3684
(b) 0.2632
(c) 0.6316

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