In a multiple choice examination there are 20 question, each question has four alternative answer following it and a student must select one correct answer. Four marks are given for the correct answer and one mark is deducted for every wrong answer. A student must score at least 50% of maximum possible marks to pass the examination. Suppose a student has not study at all so he decide to answer the question on the random bases. What is the probability that he we pass the examination?

Question

In a multiple choice examination there are 20 question, each question has four alternative answer following it and a student must select one correct answer. Four marks are given for the correct answer and one mark is deducted for every wrong answer. A student must score at least 50% of maximum possible marks to pass the examination. Suppose a student has not study at all so he decide to answer the question on the random bases. What is the probability that he we pass the examination?

Answer 1

To calculate the probability that the student will pass the examination, we need to consider two scenarios: when the student answers the questions correctly and when the student answers the questions incorrectly.

Let's calculate the probability of passing the examination by answering the questions correctly:

The total number of ways to answer the 20 questions is 4^20 (since each question has 4 possible answer choices).

The student needs to answer at least 10 questions correctly to pass the examination. The possible number of correct answers can vary from 10 to 20.

For each number of correct answers (ranging from 10 to 20), the student would also have to answer 20 - r questions incorrectly, as the remaining answers can only be incorrect.

So we need to calculate the probability for each number of correct answers and sum them up.

The probability of answering a question correctly is 1/4, and the probability of answering a question incorrectly is 3/4.

Let's calculate the probability for each scenario:

For 10 correct answers and 10 incorrect answers:
(20 choose 10) * (1/4)^10 * (3/4)^10

For 11 correct answers and 9 incorrect answers:
(20 choose 11) * (1/4)^11 * (3/4)^9

Continue this pattern up to 20 correct answers and 0 incorrect answers.

The sum of these probabilities will give us the overall probability of passing the examination.

So, the probability that the student will pass the examination by answering the questions correctly is given by:

P(pass) = (20 choose 10) * (1/4)^10 * (3/4)^10 + (20 choose 11) * (1/4)^11 * (3/4)^9 + ... + (20 choose 20) * (1/4)^20 * (3/4)^0

Next, we need to consider the probability of passing the examination by answering the questions incorrectly.

Since each question has 4 answer choices, the probability of randomly selecting the correct choice is 1/4, and the probability of randomly selecting an incorrect choice is 3/4.

The student would need to answer at least 10 questions incorrectly to pass the examination.

The probability of passing by answering the questions incorrectly avoids the need to calculate the combinations and can be expressed as:

P(fail) = (3/4)^10 + (3/4)^11 + ... + (3/4)^20

The probability of passing the examination is equal to:

P(pass) + P(fail)

We can now calculate this probability.

Answer 2

To determine the probability of passing the examination, we need to calculate the probability of scoring at least 50% of the maximum possible marks.

Each question has four alternatives, so if a student is guessing randomly, the probability of selecting the correct answer for each question is 1/4, while the probability of selecting the wrong answer is 3/4.

Let's assume the student answers all the questions randomly, without studying at all. Since there are 20 questions in total, there are 4^20 possible combinations of answers the student could select.

To calculate the probability of passing the examination, we need to consider the following possibilities:

1. The student answers all questions correctly (20 correct answers): To calculate this probability, we multiply the probability of selecting the correct answer (1/4) by the number of questions (20), so the probability is (1/4)^20.

2. The student answers 19 questions correctly and 1 question incorrectly (19 correct answers and 1 wrong answer): To calculate this probability, we need to consider all the different ways in which the student can get 19 correct answers and 1 wrong answer. There are 20 different questions that could be answered incorrectly, so the probability is 20 * (1/4)^19 * (3/4).

3. The student answers 18 questions correctly and 2 questions incorrectly (18 correct answers and 2 wrong answers): Similar to the previous case, we need to calculate the probability of getting 18 correct answers and 2 wrong answers. There are 20 * 19 / 2 different ways to choose 2 questions to be answered incorrectly, so the probability is (20 * 19 / 2) * (1/4)^18 * (3/4)^2.

We need to repeat this process for each possible combination of correct and wrong answers, up to a maximum of 10 wrong answers (since scoring more than 10 wrong answers will result in a failing grade).

Finally, we sum up all these probabilities to get the total probability of passing the examination by randomly guessing the answers.

Note: Due to the large number of calculations involved, it is not practical to perform them manually. Instead, it is recommended to use a computer program or statistical software to perform the calculations.