# Q1(a) Conditional probability from fruit counts [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-final/q1a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-final--q1a/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q1(a)** [5 marks · 2024-04-17]

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.



$$
P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\text{count satisfying A and B}}{\text{count satisfying B}}
$$



#### Solution

##### Step 1. Restrict to mangoes



$$
P(Yellow\mid Mango)=\frac{350}{650}=\frac7{13}\approx0.538462
$$



##### Step 2. Restrict to bananas



$$
P(Sweet\mid Banana)=\frac{300}{400}=\frac34=0.75
$$



##### Step 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.

**Final answer.** P(Yellow | Mango)=7/13 ≈ 0.538462; P(Sweet | Banana)=0.75.