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A number is randomly chosen from the first 100 positive inte
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30 Jul 2018, 10:40
30 Jul 2018, 10:40
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Question Stats:
81% (00:35) correct
18% (00:47) wrong based on 93 sessions
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A number is randomly chosen from the first 100 positive integers. What is the probability that it is a multiple of 3??
A. \(\frac{32}{100}\)
B. \(\frac{33}{100}\)
C. \(\frac{1}{3}\)
D. \(\frac{34}{100}\)
E. \(\frac{2}{3}\)
A. \(\frac{32}{100}\)
B. \(\frac{33}{100}\)
C. \(\frac{1}{3}\)
D. \(\frac{34}{100}\)
E. \(\frac{2}{3}\)
Re: A number is randomly chosen from the first 100 positive inte
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07 Aug 2018, 06:01
07 Aug 2018, 06:01
1
Expert Reply
Explanation
The first 100 positive integers comprise the set of numbers containing the integers 1 to 100.
Of these numbers, the only ones that are divisible by 3 are {3, 6, 9, …, 96, 99}, which adds up to exactly 33 numbers. This can be determined in several ways. One option is to count the multiples of 3, but that’s a bit slow. Alternatively, compute \(\frac{99}{3}= 33\) and realize that there are 33 multiples of 3 up to and including 99.
The number 100 is not divisible by 3, so the correct answer is \(\frac{33}{100}\).
Alternatively, use the “add one before you’re done” trick, subtracting the first multiple of 3 from the last multiple of 3, dividing by 3 and then adding 1: \(\frac{99-3}{3}+ 1 = 33\).
Then, since probability is determined by the number of desired options divided by the total number of options, the probability that the number chosen is a multiple of 3 is \(\frac{33}{100}\).
The first 100 positive integers comprise the set of numbers containing the integers 1 to 100.
Of these numbers, the only ones that are divisible by 3 are {3, 6, 9, …, 96, 99}, which adds up to exactly 33 numbers. This can be determined in several ways. One option is to count the multiples of 3, but that’s a bit slow. Alternatively, compute \(\frac{99}{3}= 33\) and realize that there are 33 multiples of 3 up to and including 99.
The number 100 is not divisible by 3, so the correct answer is \(\frac{33}{100}\).
Alternatively, use the “add one before you’re done” trick, subtracting the first multiple of 3 from the last multiple of 3, dividing by 3 and then adding 1: \(\frac{99-3}{3}+ 1 = 33\).
Then, since probability is determined by the number of desired options divided by the total number of options, the probability that the number chosen is a multiple of 3 is \(\frac{33}{100}\).
Re: A number is randomly chosen from the first 100 positive inte
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09 Aug 2018, 14:12
09 Aug 2018, 14:12
1
sandy wrote:
A number is randomly chosen from the first 100 positive integers. What is the probability that it is a multiple of 3??
A. \(\frac{32}{100}\)
B. \(\frac{33}{100}\)
C. \(\frac{1}{3}\)
D. \(\frac{34}{100}\)
E. \(\frac{2}{3}\)
A. \(\frac{32}{100}\)
B. \(\frac{33}{100}\)
C. \(\frac{1}{3}\)
D. \(\frac{34}{100}\)
E. \(\frac{2}{3}\)
Total numbers = 100
Total multiple of 3 = 33*3 = 99. It means there are 33 multiple of 3 up to 100.
Probability = 33 / 100.
The best answer is B.
