For there to be 1 car, we consider two possible outcomes:
The first door opened has a car or the second door opened has a car.
P(1 car) = 2/6 x 4/5 + 4/6 x 2/5
P(1 car) = 8/15
For there to be no car in either door
P(no car) = 4/6 x 3/5
P(no car) = 2/5
Probability of at least one car is the sum of the probability of one car and probability of two cars:
P(2 cars) = 2/6 x 1/5
= 1/15
P(1 car) + P(2 cars) = 8/15 + 1/15
= 3/5
Answer: -3 and 1/3
Step-by-step explanation:
There’s your answer
Answer:
a) A sample size of 5615 is needed.
b) 0.012
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error is:

99.5% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.
This is n for which M = 0.015.
We have that 






A sample size of 5615 is needed.
(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?
Now
.
We have to find M.



Please use " i " to denote "interest."
the formula for simple interest is i = p r t.
Here, i = $160.67 = $2000 (r) (8/12)
Solving for the interest rate, r = ($169.67)(12/8)/ $2000 = 0.127, or 12.7%