Intelligent Systems II · TEE · papers
Q1(a) Conditional probability from fruit counts [5]
Final examination · 17 Apr 2024
Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.
Consider the given dataset, apply Bayes rule and predict if a fruit has the following properties. Find (a) P(Yellow | Mango), (b) P(Sweet | Banana).
| Fruit | Yellow | Sweet | Long | Total |
|---|---|---|---|---|
| Mango | 350 | 450 | 0 | 650 |
| Banana | 400 | 300 | 350 | 400 |
| Others | 50 | 100 | 50 | 150 |
| Total | 800 | 850 | 400 | 1200 |
Answer
Given
650 mangoes, of which 350 are yellow; 400 bananas, of which 300 are sweet.
To find
The two probabilities in the printed order.
Formula and why it applies
The condition after the vertical bar restricts the denominator. For Yellow given Mango, only mangoes are eligible. This is direct conditional probability; reversing the condition would answer a different question.
LaTeX: P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\text{count satisfying A and B}}{\text{count satisfying B}}Solution
Step 1. Restrict to mangoes
LaTeX: P(Yellow\mid Mango)=\frac{350}{650}=\frac7{13}\approx0.538462Step 2. Restrict to bananas
LaTeX: P(Sweet\mid Banana)=\frac{300}{400}=\frac34=0.75Step 3. Check the direction of the condition
Using 800 in the first denominator would instead calculate P(Mango | Yellow)=350/800. Using 850 in the second would calculate P(Banana | Sweet)=300/850. Neither is the requested probability. The total 1200 cancels if you first convert every count to a joint/marginal probability.