Understanding the Exam Probability Problem

This problem involves conditional probability, where we are given information about two events and asked to find the probability of one event occurring given that the other has already occurred. Let's define the events clearly to solve this problem.

Defining Events for Probability Calculation

To accurately calculate the probability, let's assign variables to the events described in the question:

• Let E be the event that a student passes the exam.
• Let M be the event that a student gets above 90% marks in the exam.

Given Probabilities for the Exam Scenario

Based on the problem statement, we are provided with the following probabilities:

• The chance of a student passing an exam is 20%. This can be written as:

  \(P(E) = 20\% = \frac{20}{100} = 0.20\)

• The chance of a student passing the exam and getting above 90% marks in it is 5%. This means both events E and M occur simultaneously, which is their intersection:

  \(P(M \cap E) = 5\% = \frac{5}{100} = 0.05\)

Applying the Conditional Probability Formula

We are asked to find the probability that the student gets above 90% marks, given that the student passes the examination. This is a classic conditional probability question, represented as \(P(M|E)\). The formula for conditional probability is:

\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]

In our context, 'A' is getting above 90% marks (event M), and 'B' is passing the exam (event E). So, the formula becomes:

\[P(M|E) = \frac{P(M \cap E)}{P(E)}\]

Calculating the Required Probability

Now, we can substitute the given values into the conditional probability formula:

\[P(M|E) = \frac{0.05}{0.20}\]

To simplify this fraction, we can multiply the numerator and denominator by 100 to remove the decimals:

\[P(M|E) = \frac{0.05 \times 100}{0.20 \times 100} = \frac{5}{20}\]

Finally, simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 5:

\[P(M|E) = \frac{5 \div 5}{20 \div 5} = \frac{1}{4}\]

Final Probability Result for Student Marks

The probability that a student gets above 90% marks, given that the student passes the examination, is \(\frac{1}{4}\).