Suppose that in one of your classes you are given 20 questions, and are told that the final exam will consist of 8 of them. Supp
1 answer:
Answer:
a) The probability is 0.054227. You are going to be able to answer 7 questions 5.42% of the time
b) You need to be able to answer all 20 quesitons for at least a 90% chance.
Step-by-step explanation:
a) We can calculate this probability by dividing the total amount of favourable cases with the total amount of cases (because all possible exams of 8 questions are equally probable).
From 12 questions we work out on and 8 we didnt, we will be fine is at least 7 questions from the 8 given are the ones that we worked out.
The total amount of exams with exactly 7 questions we can answer is
First we pick 7 from 12, and then we pick 1 of the 8 questions we dont know the answer.
The total amount of exams with 8 questions we know the answer is the total amount of ways to pick 8 form 12, thus it is Therefore, there are a total of 6336+495 = 6831 favourable cases.
The total amount of exams, on the other hand, is the total amount of ways to pick 8 from 20, therefore it is
The probability to get 7 questions we work out to do is 6831/125970 = 0.054227. Thus, we are going to be fine only 5.42% of the time.
b) Combinatorial numbers grows pretty quickly. If we were to solve 19 out of 20 problems, then the probability of getting all 8 problems is
19/20 * 18/19 * 17/18 * 16/17 * 15/16 * 14/15 * 13/14 * 12/13 = 12/20 = 0.6
(for the first question of the exam, 19 out of 20 are fine, for the second question, of the 19 questions remaining, 18 can be answered, and so on)
This means that, if you want a 90% chance of answering all 8 questions, you need indeed be able to answer all 20 questions provided you are given.
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Answer:
a) The expected value is
b) The variance is
Step-by-step explanation:
We can assume that both marbles are withdrawn at the same time. We will define the probability as follows
#events of interest/total number of events.
We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is .
Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is . Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.
Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is . So, we define the following probabilities.
Probability of winning:
Probability of losing
Let X be the expected value of the amount you can win. Then,
E(X) = 1.10*probability of winning - 1 probability of losing =
Consider the expected value of the square of the amount you can win, Then
E(X^2) = (1.10^2)*probability of winning + probability of losing =
We will use the following formula
Thus
Var(X) =