I have a probability question that I need help with please. What is the probability that a number formed out of the ten numbers ranging 0-9 is an even number greater than 6 billion?
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Can digits be repeated?
the number is a ten-digit number with each of those numbers in it
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I'm going to suggest an answer and ask you to examine it and see if you agree.
The number must be greater than 6 billion. Of the 10 digits, the only two digits that can begin the number are 6 and 8.
There are 10 "slots" (digits) in the number. In the first slot, goes a 2 for the two ways to begin the number.
-- TWO -- -- -- -- -- -- -- -- -- --
That leaves 9 digits. An even digit has to be in the tenth slot to make the entire number even. There were 5 even digits (0,2,4,6,8) but one of those (the 6 or 8) had to begin the number. Now, there are 4 choices for the last digit.
-- TWO -- -- -- -- -- -- -- -- -- FOUR --
but also 7 and 9 can begin the number
At this point, there are 8 digits left for the remaining 8 slots of the number. For the second digit, there are 8 choices, for the third, there are 7 choices, and so on to the next to the last digit for which there will be 1 choice.
Multiply those and the number of possibilities for an even 10-digit number with value greater than 6 billion should be determined.
- TWO --EIGHT -- SEVEN -- SIX --FIVE -- FOUR -- THREE -- TWO -- ONE -- FOUR
That multiplies to 322, 560 number of ways.
Okay. The number of choices for the first digit should be changed. What would you put there? And, what would be place as the last digit?
Would you have to do two different probabilities because 2 chances of odd then 5 chances of even at the end or 2 chances even then 4 chances even?
I suppose the number of favorable outcomes could be calculated that way. What did you get?
The site went down. I cranked out the problem using two cases and wondered if you had had time to do that?