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2, 3, and 5 are substituted at random for a, b, and c in the equation ax+b=c. Probability that solution is negative? P( absolute value = 1?

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ok, well you can brute force this
if assignment is random, then the probability of something is always (number of success possibilities)/ (total number of possibilities)
so there are a total of six possibilities and we can actually list them all:

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hmmm actually ok misread, there is an x variable haha
ok so let's list the six possibilities anyway and see where that gets us
1. 2x + 3 = 5 2. 2x + 5 = 3 3. 3x + 5 = 2 4. 3x + 2 = 5 5. 5x + 3 = 2 6. 5x + 2 = 3
k. im with you so far
ok cool, so now let's solve the cases
1. 2x = 2. x = 1 (not negative)
2. 2x = -2. x = -1 (negative)
3. 3x = -3. x = -1. (negative)
4. 3x = 3. x = 1. (positive)
5. 5x = -1. x = -1/5. (negative)
6. 5x = 1. x = 1/5 (positive)
so out of the six possibilities, 3 are positive, 3 are negative
so thats a 50 percent probabilty
and sometimes there's no systematic theorem or rule - it's just #outcomes/#possible outcomes

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