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how ya doing?
Fine until I saw this problem. :) Just kidding, CW.
Ha i have tons of them....
Find out how many different numbers you can make out of the three digits, and how many of those are less than 700. There are 6*5*4=120 ways to make a three digit number out of the 6 given digits. Out of these, if we were to make a number less than 700, the first digit must be less than 7 (three choices), the remaining digits have 5 and 4 choices for a total of 3*5*4=60 ways. So probability P(<700)=N(<700)/N(all)
so is it 60/700
N(all) is 6*5*4=120 ways with no restrictions.
what would the probability be then?
So probability P(<700)=N(<700)/N(all)
im still not sure how this answer is supposed to look
Any clues directrix?
There are 60 (desirable) ways to make 3-digit numbers less than 700 out of 120 (possible) ways to make 3-digit numbers out of the given 6 distinct digits. So the probability required, P(<700) is given by P(<700) = number of desirable events / number of possible events.
We are making 3-digit numbers that are less than 700. So, the first digit of the 3-digit number cannot be 7, 8, or 9. For the first digit, we have 3 choices (1,2, or 3). After selecting one of (1,2,3) for the first digit, there are 5 choices left for the number's second digit, and then 4 choices for the third digit. By the Fundamental Principle of Counting, there are 3 times 5 times 4 = 60 ways to make a 3-digit number less than 700 from the given digits. 60 is the desired number of outcomes. For the probability denominator, we need the number of possible outcomes of the formation of the 3-digit number. That would be 6 times 5 times 4 or 120. The probability of creating a 3-digit number less than 700 from the digits 1,2,3,7,8,9 is 60/120 or 1/2.
Sorry, CW, I had to stop eat, and then the server threw me off. I'm back now.
CW, did you see my work here? Do you agree?