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
Step-by-step explanation:
Here are some examples of equations equal to 32.
16x2=32
64/2=32

- a number

- its square

- the sum

- less than or equal
negative three - I don't have to explain this one, do I?

- the sum of a number and its square is less than or equal negative three
Answer:
Step-by-step explanation:
Sample proportion p is the proportion of favourable numbers to total number in the sample
By central limit theorem and also approximation of binomial to normal , we have sample proportion for large number of samples will be normal
with mean = sample proportion
and std deviation = 
Thus we find standard deviation of proportion sample is inversely proportional to the square of the sample size n.
It follows automatically that as sample size increases std deviation decreases.
Here from 80 sample size was made to 200
So std deviation would decrease automatically