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03 Mar 2004, 15:51
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HOw many different six digit #s exist, the sum of whose digits is odd?
5*10^5
4*10^5
45*10^4
9*10^4
5*10^5
4*10^5
45*10^4
9*10^4
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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03 Mar 2004, 16:18
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I'd say 45 * 10^4
Total outcome: 9 * 10^5
Explanation: 9 is number of ways first digit can be selected(obviously, 0 cannot be the first number of the 6-digit number so there is only 9 possible choices)
10^5 is number of ways to select each of the remaining 5 digits
What you need now in order to have their sum being odd is:
E+E+E+E+E+O
E+E+E+O+O+O
E+O+O+O+O+O
As can be seen, you need either 1 odd, 3 odd or 5 odd numbers in order to have the sum being odd. However, the odd of having 1 even, 3 even or 5 even numbers is just the same. Therefore, probability of having odd/(total odd+even) is 1/2
Answer is 1/2*9*10^5 = 45*10^4
Total outcome: 9 * 10^5
Explanation: 9 is number of ways first digit can be selected(obviously, 0 cannot be the first number of the 6-digit number so there is only 9 possible choices)
10^5 is number of ways to select each of the remaining 5 digits
What you need now in order to have their sum being odd is:
E+E+E+E+E+O
E+E+E+O+O+O
E+O+O+O+O+O
As can be seen, you need either 1 odd, 3 odd or 5 odd numbers in order to have the sum being odd. However, the odd of having 1 even, 3 even or 5 even numbers is just the same. Therefore, probability of having odd/(total odd+even) is 1/2
Answer is 1/2*9*10^5 = 45*10^4
03 Mar 2004, 16:22
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Paul,
I appreciate your explanation. Thanx!
I appreciate your explanation. Thanx!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
