A test consists of 5 multiple questions each of which 3 answers are given, only 1 of which is correct.for each correct answer a candidate gets 2 marks but loses 1 mark for each incorrect answer.draw up table to show all possible marks and a probability associated with each mark for these candidates.
Cannot do table on these posts.
To draw up a table showing all possible marks and the probability associated with each mark, we need to consider the different combinations of correct and incorrect answers. We can use the concept of the binomial coefficient to find the number of ways each combination can occur.
Here's how we can approach this problem step by step:
Step 1: Determine the number of correct answers
Since there are 5 questions and 3 possible answers for each question, the number of correct answers can be any value from 0 to 5.
Step 2: Calculate the number of ways to choose the correct answers
To calculate the number of ways to choose the correct answers, we can use the binomial coefficient formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of questions and k is the number of correct answers.
For example, if we want to find the number of ways to choose 3 correct answers out of 5 questions, the calculation would be: C(5, 3) = 5! / (3!(5-3)!) = 10.
Step 3: Calculate the number of ways to choose the incorrect answers
The number of incorrect answers will be the difference between the total number of questions and the number of correct answers.
Step 4: Calculate the probability for each mark
Since each correct answer is worth 2 marks and each incorrect answer results in a loss of 1 mark, we can determine the total marks for each combination. For example, if there are 3 correct answers and 2 incorrect answers, the total marks would be: (3 * 2) + (2 * (-1)) = 6 - 2 = 4.
Finally, to determine the probability associated with each mark, we need to divide the number of ways each combination can occur by the total number of possible combinations, which is 3^5 or 243 in this case.
By following these steps for each possible number of correct answers, we can construct the table showing all possible marks and their corresponding probabilities.
Here is an example of such a table:
Number of Correct Answers | Number of Incorrect Answers | Total Marks | Probability
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0 | 5 | -5 | x
1 | 4 | -3 | y
2 | 3 | 1 | z
3 | 2 | 5 | w
4 | 1 | 9 | v
5 | 0 | 13 | u
Note that the probabilities x, y, z, w, v, and u can be calculated by dividing the number of ways each combination can occur by 243, which is the total number of possible combinations in this case.