Mathematics OpenStudy (anonymous):

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? OpenStudy (anonymous):

ok, well you can brute force this OpenStudy (anonymous):

if assignment is random, then the probability of something is always (number of success possibilities)/ (total number of possibilities) OpenStudy (anonymous):

so there are a total of six possibilities and we can actually list them all: OpenStudy (anonymous):

hmmm actually ok misread, there is an x variable haha OpenStudy (anonymous):

ok so let's list the six possibilities anyway and see where that gets us OpenStudy (anonymous):

1. 2x + 3 = 5 2. 2x + 5 = 3 3. 3x + 5 = 2 4. 3x + 2 = 5 5. 5x + 3 = 2 6. 5x + 2 = 3 OpenStudy (anonymous):

k. im with you so far OpenStudy (anonymous):

ok cool, so now let's solve the cases OpenStudy (anonymous):

1. 2x = 2. x = 1 (not negative) OpenStudy (anonymous):

2. 2x = -2. x = -1 (negative) OpenStudy (anonymous):

3. 3x = -3. x = -1. (negative) OpenStudy (anonymous):

4. 3x = 3. x = 1. (positive) OpenStudy (anonymous):

5. 5x = -1. x = -1/5. (negative) OpenStudy (anonymous):

6. 5x = 1. x = 1/5 (positive) OpenStudy (anonymous):

so out of the six possibilities, 3 are positive, 3 are negative OpenStudy (anonymous):

so thats a 50 percent probabilty OpenStudy (anonymous):

exactly OpenStudy (anonymous):

and sometimes there's no systematic theorem or rule - it's just #outcomes/#possible outcomes OpenStudy (anonymous):

right.

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