Answer:
Most people found the probability of just stopping at the first light and the probability of just stopping at the second light and added them together. I'm just going to show another valid way to solve this problem. You can solve these kinds of problems whichever way you prefer.
There are three possibilities we need to consider:
Being stopped at both lights
Being stopped at neither light
Being stopped at exactly one light
The sum of the probabilities of all of the events has to be 1 because there is a 100% chance that one of these possibilities has to occur, so the probability of being stopped at exactly one light is 1 minus the probability of being stopped at both lights minus the probability of being stopped at neither.
Because the lights are independent, the probability of being stopped at both lights is just the probability of being stopped at the first light times the probability of being stopped at the second light. (0.4)(0.7) = 0.28
The probability of being stopped at neither is the probability of not being stopped at the first light, which is 1-0.4 or 0.6, times the probability of not being stopped at the second light, which is 1-0.7 or 0.3. (0.6)(0.3) = 0.18
The probability at being stopped at exactly one light is 1-0.18-0.28=.54 or 54%.
Let x be the first number
X + 1 be the second number
So
(x+1)^2 – x^2 = 2017
X^2 + 2x +1 – x^2 = 2017
2x + 1 = 2017
2x = 2017 – 1
2x = 2016
X = 1008
X + 1 = 1009
Sum = 1008 + 1009 = 2017
