Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.
For each pill, there are only two possible outcomes, either it is acceptable, or it is not. The probability of a pill being acceptable is independent of any other pill, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- The sample has 20 pills, hence
.
- 100 - 4 = 96% are acceptable, hence

The probability that <u>fewer that 5 in a sample of 20 pills</u> will be acceptable is:

In which






0% probability that fewer that 5 in a sample of 20 pills will be acceptable.
A similar problem is given at brainly.com/question/24863377
So, since the student makes $30 each week, we can disregard the number of hours.
So, we can model this situation by the following equation
9.5h+30
So, if he works 10.25 hrs (15 minutes = .25 of 1 hr) we just plug that number in

Since it's money we round to the nearest cent
So he will make $127.28 Next week
Answer:
23.44%
Step-by-step explanation:
The probability of getting a 4 on the first 2 throws and different numbers on the last 5 throws = 1/6 * 1/6 * (5/6)^5
= 0.01116
There are 7C2 ways of the 2 4's being in different positions
= 7*6 / 2 = 21 ways.
So the required probability = 0.01116 * 21
= 0.2344 or 23.44%.