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A three-digit positive integer is chosen at random. What is the probab
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04 Jun 2021, 23:20
04 Jun 2021, 23:20
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A three-digit positive integer is chosen at random. What is the probability that the product of its digits is even?
A. 1/2
B. 31/36
C. 49/54
D. 7/8
E. 11/12
A. 1/2
B. 31/36
C. 49/54
D. 7/8
E. 11/12
Re: A three-digit positive integer is chosen at random. What is the probab
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06 Jun 2021, 05:58
06 Jun 2021, 05:58
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Solution:
There are total 900 three digit integers.
Now, to find an integer whose product is not even, neither of the 3 digits in the integer should be even as e*o*o=e
Thus, all the three digits has to be even and this can be done in 5* 5* 5=125 ways
Therefore, probability that the product of the digit is even=\(\frac{900-125}{900}\)=\(\frac{31}{35}\)
IMO B
Hope this helps!
There are total 900 three digit integers.
Now, to find an integer whose product is not even, neither of the 3 digits in the integer should be even as e*o*o=e
Thus, all the three digits has to be even and this can be done in 5* 5* 5=125 ways
Therefore, probability that the product of the digit is even=\(\frac{900-125}{900}\)=\(\frac{31}{35}\)
IMO B
Hope this helps!
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Re: A three-digit positive integer is chosen at random. What is the probab
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06 Jun 2021, 09:53
06 Jun 2021, 09:53
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Total Even numbers can be found by taking even digit in any of the three places. So there will be lot of combinations and it will be difficult to list down everything.
This is why an approach to find odd number is easier and will save you time.
If you want to find total number of even numbers then they this.
First digit even (4 ways) * anything in second and third. So 400 ways.
Second digit even and first digit odd and anything in third place.
So, 5*5*10 = 250 ways
Third digit even and first and second are odd = 5*5*5 = 125 ways
Total = 400+250+125 = 775.
Hope it helps!
This is why an approach to find odd number is easier and will save you time.
If you want to find total number of even numbers then they this.
First digit even (4 ways) * anything in second and third. So 400 ways.
Second digit even and first digit odd and anything in third place.
So, 5*5*10 = 250 ways
Third digit even and first and second are odd = 5*5*5 = 125 ways
Total = 400+250+125 = 775.
Hope it helps!
void wrote:
why total even no. can be 9*10*5
9= from 1 to 9 for hundred position
10=for 10 choice tenth digit position
5= 0,2,4,6,8 possible option unit place
9= from 1 to 9 for hundred position
10=for 10 choice tenth digit position
5= 0,2,4,6,8 possible option unit place
