# Q1(b) Bayes posterior for red ball [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-reexam/q1b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-reexam--q1b/

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(b)** [5 marks · 2024-06-25]

Bag A contains 3 red and 4 green balls. Bag B contains 4 red and 5 green balls. One ball is drawn at random from one of the bags and found red. Find the probability it came from Bag A.

### Answer

#### Given

A has 7 balls, B has 9; red likelihoods 3/7 and 4/9.

#### To find

P(A|red).

#### Formula and why it applies

Reverse the direction of the evidence using Bayes. The probability of choosing each bag is a prior; the probability of red inside a chosen bag is a likelihood.



$$
P(A\mid R)=\frac{P(R\mid A)P(A)}{P(R\mid A)P(A)+P(R\mid B)P(B)}
$$



#### Solution

**Source / assumption note.** Assume a bag is chosen uniformly, then a ball uniformly within that bag. Choosing a ball uniformly from all 16 balls would be a different experiment.

##### Step 1. Weight each red likelihood by its bag prior



$$
P(R,A)=\frac37\frac12=\frac3{14},\quad P(R,B)=\frac49\frac12=\frac29
$$



##### Step 2. Add all ways to see red



$$
P(R)=\frac3{14}+\frac29=\frac{27+28}{126}=\frac{55}{126}
$$



##### Step 3. Normalize the A contribution



$$
P(A\mid R)=\frac{3/14}{55/126}=\frac{27}{55}\approx0.490909
$$



**Final answer.** 27/55 ≈ 0.490909 under equal bag-selection priors.