Exercise Type 7: Ball/Box Conditional Probability

What the exam asks: You have boxes/bags with different objects (apples, oranges, red balls, green balls). You pick a box, draw an object, and must compute probabilities. Sometimes objects are NOT returned (without replacement), which changes the counts.

Part 0: What Do All These Symbols Mean?

The Key Notation

Symbol	How to Read It	What It Means
$P(A)$	"probability of A"	How likely event A is
$P(A\|B)$	"probability of A given B"	How likely A is IF we know B already happened
$P(A, B)$	"probability of A and B"	How likely both A and B happen together
$P(B_1)$	"probability of box 1"	How likely we pick box 1
$P(\text{apple}\|B_1)$	"probability of apple given box 1"	If we picked box 1, what's the chance we draw an apple?
$P(B_1\|\text{apple})$	"probability of box 1 given apple"	We drew an apple — what's the chance it came from box 1?

The Core Concepts

1. The Product Rule:

$P(A, B) = P(A|B) \cdot P(B)$

In plain English: "Probability of both A and B happening = (probability of A if B happened) × (probability of B)"

Example: "Probability of picking box 1 AND drawing an apple = (probability of apple if box 1) × (probability of box 1)"

2. Law of Total Probability:

$P(A) = P(A|B_1) \cdot P(B_1) + P(A|B_2) \cdot P(B_2)$

In plain English: "Total probability of A = (probability of A in scenario 1 × probability of scenario 1) + (probability of A in scenario 2 × probability of scenario 2)"

3. Bayes' Rule:

$P(B_1|A) = \frac{P(A|B_1) \cdot P(B_1)}{P(A)}$

In plain English: "Probability that scenario 1 was the cause, given we saw A = (how likely scenario 1 produces A × how likely scenario 1 is) ÷ (total probability of A)"

Part 1: FULL Walkthrough of Real Exam Questions

EXAM QUESTION 1 (2021-Resit, Questions 1a-1b)

Box 1: 4 apples, 8 oranges (12 total) Box 2: 10 apples, 2 oranges (12 total) Boxes chosen with equal probability (50/50) One draw.

Question 1a: What is the probability of choosing an apple?

STEP-BY-STEP SOLUTION

Step 1: Understand what we're computing

We want $P(\text{apple})$. We don't know which box we'll pick, so we need to consider both possibilities.

Step 2: Use the Law of Total Probability

$P(\text{apple}) = P(\text{apple}|B_1) \cdot P(B_1) + P(\text{apple}|B_2) \cdot P(B_2)$

Step 3: Fill in each piece

$P(B_1) = 1/2$ (equal probability)
$P(B_2) = 1/2$ (equal probability)
$P(\text{apple}|B_1) = 4/12$ (4 apples out of 12 total in box 1)
$P(\text{apple}|B_2) = 10/12$ (10 apples out of 12 total in box 2)

Step 4: Compute

$P(\text{apple}) = \frac{4}{12} \cdot \frac{1}{2} + \frac{10}{12} \cdot \frac{1}{2} = \frac{4}{24} + \frac{10}{24} = \frac{14}{24} = \frac{7}{12}$

Answer: 7/12 ✅

Question 1b: If an apple is chosen, what is the probability it came from Box 1?

STEP-BY-STEP SOLUTION

Step 1: Understand what we're computing

We want $P(B_1|\text{apple})$. This is the REVERSE of what we computed before — now we KNOW we got an apple, and we want to know which box it likely came from.

This is Bayes' rule!

Step 2: Write Bayes' rule

$P(B_1|\text{apple}) = \frac{P(\text{apple}|B_1) \cdot P(B_1)}{P(\text{apple})}$

Step 3: Plug in the numbers

$P(\text{apple}|B_1) = 4/12$
$P(B_1) = 1/2$
$P(\text{apple}) = 7/12$ (from part a)

$P(B_1|\text{apple}) = \frac{\frac{4}{12} \cdot \frac{1}{2}}{\frac{7}{12}} = \frac{4/24}{7/12}$

Step 4: Simplify the fraction

$\frac{4/24}{7/12} = \frac{4}{24} \cdot \frac{12}{7} = \frac{4 \cdot 12}{24 \cdot 7} = \frac{48}{168} = \frac{4}{14} = \frac{2}{7}$

Note: The exam listed 1/3 as an option, but the correct answer depends on the exact numbers. The method is what matters.

EXAM QUESTION 2 (2023, Question 4d — Without Replacement)

A dark bag contains 5 red balls and 7 green balls (12 total). Balls are NOT returned after each draw. You know that on the LAST draw, the ball was green. What is the probability of drawing a red ball on the FIRST draw?

Options: (a) $4/11$, (b) $5/11$, (c) $5/12$, (d) $6/11$

STEP-BY-STEP SOLUTION

Step 1: Understand the scenario

We draw balls one at a time without putting them back. We're told the LAST ball was green. Given this information, what's the probability the FIRST ball was red?

We want: $P(R_1 | G_{\text{last}})$

Step 2: Write Bayes' rule

$P(R_1 | G_{\text{last}}) = \frac{P(G_{\text{last}} | R_1) \cdot P(R_1)}{P(G_{\text{last}})}$

Step 3: Compute $P(R_1)$

Before any draws, the bag has 5 red and 7 green:

$P(R_1) = \frac{5}{12}$

Step 4: Compute $P(G_{\text{last}} | R_1)$

If the first ball was RED, the remaining balls are: - 4 red, 7 green (11 total)

So the probability the last ball is green (from the remaining 11):

$P(G_{\text{last}} | R_1) = \frac{7}{11}$

Step 5: Compute $P(G_{\text{last}})$

Using the Law of Total Probability:

$P(G_{\text{last}}) = P(G_{\text{last}} | R_1) \cdot P(R_1) + P(G_{\text{last}} | G_1) \cdot P(G_1)$

$P(R_1) = 5/12$
$P(G_1) = 7/12$
$P(G_{\text{last}} | R_1) = 7/11$ (if first was red: 4 red, 7 green remain)
$P(G_{\text{last}} | G_1) = 6/11$ (if first was green: 5 red, 6 green remain)

$P(G_{\text{last}}) = \frac{7}{11} \cdot \frac{5}{12} + \frac{6}{11} \cdot \frac{7}{12} = \frac{35}{132} + \frac{42}{132} = \frac{77}{132} = \frac{7}{12}$

Important insight: By symmetry, $P(G_{\text{last}}) = P(G_1) = 7/12$. The marginal probability of green is the same regardless of position!

Step 6: Apply Bayes' rule

$P(R_1 | G_{\text{last}}) = \frac{\frac{7}{11} \cdot \frac{5}{12}}{\frac{7}{12}} = \frac{7}{11} \cdot \frac{5}{12} \cdot \frac{12}{7} = \frac{5}{11}$

The $7/12$ cancels with $12/7$, and the 7s cancel.

Answer: (b) 5/11 ✅

Part 2: Tricks & Shortcuts

TRICK 1: Symmetry for Marginal Probabilities

$P(G_{\text{last}}) = P(G_1)$ when you don't know intermediate draws. The marginal probability doesn't depend on position.

TRICK 2: "Without Replacement" Means Update Counts

After drawing a red ball: - Red count decreases by 1 - Total count decreases by 1

TRICK 3: Future Conditions Past

$P(\text{first} | \text{last})$ uses Bayes' rule with the future as "evidence."

TRICK 4: Tree Diagram Helps

Draw all possible sequences as branches. Multiply along branches, add across paths.

Part 3: Practice Exercises

Exercise 1

Box 1: 4 apples, 8 oranges. Box 2: 10 apples, 2 oranges. Boxes chosen with equal probability. One draw.

What is the probability of choosing an apple?

Exercise 2

Same setup. If an apple is chosen, what is the probability it came from Box 1?

Exercise 3

A dark bag contains 5 red balls and 7 green ones. Balls are not returned. If you know that on the last draw the ball was green, what is the probability of drawing a red ball on the first draw?

Options: - (a) $4/11$ - (b) $5/11$ - (c) $5/12$ - (d) $6/11$

Answers

Exercise 1

**Answer: 7/12** P(apple) = P(apple|B₁)P(B₁) + P(apple|B₂)P(B₂) = (4/12)(1/2) + (10/12)(1/2) = 4/24 + 10/24 = 14/24 = 7/12

Exercise 2

**Answer: 2/7** P(B₁|apple) = P(apple|B₁)P(B₁) / P(apple) = (4/12 × 1/2) / (7/12) = (4/24) / (7/12) = 4/24 × 12/7 = 48/168 = 2/7

Exercise 3

**Answer: (b) 5/11** P(R₁|G_last) = P(G_last|R₁)P(R₁) / P(G_last) = (7/11 × 5/12) / (7/12) = 5/11