Answer:

Step-by-step explanation:
<u>Probabilities</u>
The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.
Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

Thus the option WN has the probability

Now for the second option NW. The initial probability to pick a non-white counter is

The probability to pick a white counter is

Thus the option NW has the probability

The total probability of event A is the sum of both


Two multiplied by x plus three plus six is equal to three multiplied by x subtracted by nine
Answer:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
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 z-score that has a p-value of
.
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.
This means that 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).