Sam=3syd+6 and sam=36, since sam=sam we can say:
3syd+6=36
3syd=30
syd=10
So Sydney has 10 pieces of candy.
Looking at the set, we are given 18 elements. 17 is prime; it has only two factors: 1 and 17, since 1•17=17. So, the question is really asking what is the probability the numbers 1 or 17 is chosen. As mentioned earlier, 17 is prime, so there are two possible choices: 1 and 17.
P (probability) = possible outcomes / total outcomes
It is important to note that these events are “or” events, meaning that the probability can only be determined by choosing a 1 or a 17; you can’t randomly chose a 1 and 17 at the same time. So, the formula is:
P(A or B) = P(A) + P(B)
All this is saying is that given two possible outcomes, the probability occurs independent of each event; they don’t occur at the same time.
P(1 or 17) = P(1)/18 + P(1)/18
P(1 or 17) = 2/18
Since 17 is prime, it’s two and only factors are 1 and 17. The probability of randomly choosing a 1 or 17 is 2/18, meaning that there are 2 elements in the set out of a possible 18 elements that can be randomly chosen.
2/18 simplifies to 1/9
So, your answer is 1/9
