The two last subparts please (d and e) 6. An integer k E {1, 2, ..., 999, 1000} is selected at random. What is the probabilitythat:(a) k is divisible by 3?(b) k is divisible by 4?(c) 2k is divisible by 3?(d) k is divisible by 5 but not by 3?(e) k is a prime number or a composite number with a prime factor p < 29?

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Description: The two last subparts please (d and e) 6. An integer k E {1, 2, ..., 999, 1000} is selected at random. What is the probabilitythat:(a) k is divisible by 3?(b) k is divisible by 4?(c) 2k is divisible by 3?(d) k is divisible by 5 but not by 3?(e) k is

Answered: The two last subparts please (d and e

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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The two last subparts please (d and e)

6. An integer k E {1, 2, ..., 999, 1000} is selected at random. What is the probability
that:
(a) k is divisible by 3?
(b) k is divisible by 4?
(c) 2k is divisible by 3?
(d) k is divisible by 5 but not by 3?
(e) k is a prime number or a composite number with a prime factor p < 29?

Expert Solution

Step 1

(d)

The number of numbers from 1 to 1000 divisible by 5 is 1000/5=200. If we want to count the number of numbers out of these, that are not divisible by 3, then we must remove those numbers that are divisible by both 3 and 5, those will be all the numbers that are divisible by 3*5=15, the number of such numbers is 1000/15=66. So, the answer is 200–66=134

Total numbers less than 1000 which are divisible by 3 = $[1000 / 3] = 333$

Total numbers less than 1000 which are divisible by 15 = $[1000 / 15] = 66$

So total numbers less than 1000 which are divisible by 5 but not by 3 =200 $333 - 66 = 267$

$333 - 66 = 267$

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