Question

Find the probability that a student selected at random is a regular student.

Answer 1

Concept:
• The sum of the probabilities of all mutually exclusive and exhaustive events is 1.
• If an event \(D\) (dropouts) has a probability \(P(D)\), then its complement \(R\) (regular students) has probability \(P(R) = 1 - P(D)\).

Step 1: Identify the given dropout percentage
Given that the percentage of dropout students is \(40\%\). So, \(P(\text{Dropout}) = \frac{40}{100} = 0.4\).

Step 2: Calculate the probability of regular students
The remaining students are regular students: \[ P(\text{Regular}) = 1 - P(\text{Dropout}) \] \[ P(\text{Regular}) = 1 - 0.4 = 0.6 \] Alternatively, this is \(60\%\).

Answer 2

Concept:

• Conditional probability \(P(F|D)\) is the probability of failure given the student is a dropout.
• \(P(\text{Qualify}) + P(\text{Not Qualify}) = 1\) for any specific group.

Step 1: Identify the success rate for dropouts
Given that for the dropout group, the qualifying rate (success rate) is \(5\%\).
\[ P(\text{Qualify}|\text{Dropout}) = 0.05 \]

Step 2: Calculate the non-qualifying probability
The probability of not qualifying is the complement:
\[ P(\text{Not Qualify}|\text{Dropout}) = 1 - 0.05 \] \[ P(\text{Not Qualify}|\text{Dropout}) = 0.95 \]

Answer 3

Concept:
• Use Bayes' Theorem to find the probability of a cause given an effect.
• \(P(\text{Regular}|\text{Qualify}) = \frac{P(\text{Qualify}|\text{Regular}) \cdot P(\text{Regular})}{P(\text{Total Qualify})}\).
• Total Probability \(P(Q) = P(Q|D)P(D) + P(Q|R)P(R)\).

Step 1: Calculate the total probability of qualifying
Let \(Q\) be qualifying, \(D\) be dropout, \(R\) be regular. \(P(D) = 0.4, P(R) = 0.6\) \(P(Q|D) = 0.05, P(Q|R) = 0.10\) \[ P(Q) = (0.05)(0.4) + (0.10)(0.6) \] \[ P(Q) = 0.02 + 0.06 = 0.08 \]

Step 2: Apply Bayes' Theorem
The student is "not a dropout", meaning they are a regular student. We need \(P(R|Q)\): \[ P(R|Q) = \frac{P(Q|R)P(R)}{P(Q)} \] \[ P(R|Q) = \frac{0.10 \times 0.6}{0.08} = \frac{0.06}{0.08} \]

Step 3: Simplify the final value
\[ P(R|Q) = \frac{6}{8} = \frac{3}{4} = 0.75 \]

Answer 4

Concept:

• This is also a Bayes' Theorem application based on failure probabilities.
• Total Failure Probability \(P(F) = 1 - P(Q)\).

Step 1: Calculate the failure probabilities for each group
Let \(F\) be failure (not qualifying).
\(P(F|D) = 1 - 0.05 = 0.95\) \(P(F|R) = 1 - 0.10 = 0.90\)

Step 2: Find the total probability of failure
Using the result from the previous part \(P(Q) = 0.08\):
\[ P(F) = 1 - 0.08 = 0.92 \]

Step 3: Apply Bayes' Theorem for \(P(R|F)\)
\[ P(R|F) = \frac{P(F|R)P(R)}{P(F)} \] \[ P(R|F) = \frac{0.90 \times 0.6}{0.92} = \frac{0.54}{0.92} \]

Step 4: Simplify the fraction
\[ P(R|F) = \frac{54}{92} = \frac{27}{46} \]