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25 Dec 2017, 01:44
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Difficulty:
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58% (01:55) correct
42%
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What are the odds that a 3-digit number contains 3 distinct prime numbers?
A. 3/125
B. 2/75
C. 24/899
D. 3/50
E. 1/15
A. 3/125
B. 2/75
C. 24/899
D. 3/50
E. 1/15
25 Dec 2017, 02:47
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Bunuel
What are the odds that a 3-digit number contains 3 distinct prime numbers?
A. 3/125
B. 2/75
C. 24/899
D. 3/50
E. 1/15
A. 3/125
B. 2/75
C. 24/899
D. 3/50
E. 1/15
The tricky part of this question is understanding the question...
We'll break it down, a Logical approach.
If a '3-digit number contains 3 distinct prime numbers' then each of its digits must be a different prime number.
So, we need to know how many single digit prime numbers there are.
These are 2,3,5, and 7.
Then there are 4*3*2 total ways to arrange 3 of these numbers.
Since we have a total of 9 options for the hundreds digit (it cannot be 0!) and 10 options for the tens and ones digits, there are a total of 9*10*10 3 digit numbers.
This gives odds of (4*3*2)/(9*10*10) = 24/900 = 12/450 = 6/225 = 2/75
(B) is our answer.
25 Dec 2017, 16:37
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Bunuel
What are the odds that a 3-digit number contains 3 distinct prime numbers?
A. 3/125
B. 2/75
C. 24/899
D. 3/50
E. 1/15
A. 3/125
B. 2/75
C. 24/899
D. 3/50
E. 1/15
One-digit distinct prime numbers are {2,3,5,7}.
From 999 to 100 their are \(999-100+1=900\) different 3-digit numbers.
Favorable outcomes starting from 200 is: {235,237,253,257,273,257} = 6, and the same result will also be for, 300, 500 and 700.
So, the favorable outcomes for all 3-digit number containing 3 distinct prime numbers is \(6*4=24\).
Thus, odds are: \(\frac{Favorable.Outcomes}{Total.Outcomes}=\frac{24}{900}=8/300=4/150=2/75\)
(B) is the answer.
