Probability with Independent Events: Fruit Selection

Probability 9th-10th Grade
PROBLEM
Carol purchased one basket of fruit consisting of 4 apples and 2 oranges and another basket of fruit consisting of 3 apples and 5 oranges. Carol is to select one piece of fruit at random from each of the two baskets. What is the probability that one of the two pieces of fruit selected will be an apple and the other will be an orange?

What This Problem Teaches

  • Breaking complex probability problems into mutually exclusive cases
  • Distinguishing between independent events (selections from different baskets) and mutually exclusive events (different successful outcomes)
  • Applying the multiplication rule for AND events and addition rule for OR events
  • Converting word problems into precise mathematical probability statements
  • Verifying probability answers using complementary counting

Let's Draw It

Before we calculate, let's visualize what we're working with. This tree diagram shows all possible outcomes when selecting one fruit from each basket.

Carol purchased one basket of fruit consisting of 4 apples and 2 oranges and another basket of fruit consisting of 3...

Solution: Method 1 — Case Analysis with Tree Diagram Thinking

The key insight is recognizing that "one apple and one orange" can happen in exactly two mutually exclusive ways. Let's work through each case systematically.

Step 1 — Identify the successful outcomes

We want one apple and one orange from our two selections. This can happen in two ways:

  • Case A: Apple from Basket 1 and Orange from Basket 2
  • Case B: Orange from Basket 1 and Apple from Basket 2

Step 2 — Calculate the probability for Case A

Basket 1 has 4 apples out of 6 total fruits, so P(Apple from Basket 1) = 4/6 = 2/3

Basket 2 has 5 oranges out of 8 total fruits, so P(Orange from Basket 2) = 5/8

P(Case A) = P(Apple from B1) × P(Orange from B2)
P(Case A) = (4/6) × (5/8) = 20/48

Step 3 — Calculate the probability for Case B

Basket 1 has 2 oranges out of 6 total fruits, so P(Orange from Basket 1) = 2/6 = 1/3

Basket 2 has 3 apples out of 8 total fruits, so P(Apple from Basket 2) = 3/8

P(Case B) = P(Orange from B1) × P(Apple from B2)
P(Case B) = (2/6) × (3/8) = 6/48

Step 4 — Add the probabilities of the two cases

Since Case A and Case B are mutually exclusive (they can't happen at the same time), we add their probabilities:

P(one apple and one orange) = P(Case A) + P(Case B)
P(one apple and one orange) = 20/48 + 6/48 = 26/48

Step 5 — Simplify the fraction

We can reduce 26/48 by finding the greatest common factor:

26/48 = 13/24

Solution: Method 2 — Complementary Counting with All Outcomes

Instead of focusing on successful cases, let's count all possible outcomes and identify which ones give us one apple and one orange.

Step 1 — Set up the outcome matrix

There are 6 × 8 = 48 total possible outcomes. Let's organize them by what we select from each basket:

Basket 1 → Basket 2Apple (3)Orange (5)
Apple (4)4 × 3 = 12 outcomes
(Both apples)		4 × 5 = 20 outcomes
✓ SUCCESS
Orange (2)2 × 3 = 6 outcomes
✓ SUCCESS		2 × 5 = 10 outcomes
(Both oranges)

Step 2 — Count successful outcomes

The successful outcomes (one apple and one orange) are:

  • Apple from Basket 1, Orange from Basket 2: 20 outcomes
  • Orange from Basket 1, Apple from Basket 2: 6 outcomes

Total successful outcomes: 20 + 6 = 26

Step 3 — Calculate the probability

P(one apple and one orange) = Successful outcomes / Total outcomes
P(one apple and one orange) = 26/48 = 13/24
The probability is 13/24 ≈ 0.542 or about 54.2%

Verification

Let's verify our answer using two different approaches:

Check 1 — Verify all probabilities sum to 1

All possible outcomes should have probabilities that sum to 1:

  • Both apples: (4/6) × (3/8) = 12/48 = 1/4
  • One apple, one orange: 26/48 = 13/24
  • Both oranges: (2/6) × (5/8) = 10/48 = 5/24
1/4 + 13/24 + 5/24 = 6/24 + 13/24 + 5/24 = 24/24 = 1 ✓

Check 2 — Direct calculation verification

Using our case analysis:

Case A: (4/6) × (5/8) = 20/48
Case B: (2/6) × (3/8) = 6/48
Total: 20/48 + 6/48 = 26/48 = 13/24 ✓

Watch Out for These

✗ Mistake 1: Adding instead of multiplying within each case

Incorrect: P(Apple from B1 and Orange from B2) = 4/6 + 5/8
This treats independent events as mutually exclusive. When you want both events to happen (apple AND orange), you multiply their probabilities.

✗ Mistake 2: Forgetting one of the two successful cases

Incorrect: Only calculating (4/6) × (5/8) = 20/48
This misses the case where we get orange from Basket 1 and apple from Basket 2. The problem asks for "one apple and one orange" — it doesn't matter which basket each comes from.

✗ Mistake 3: Using the wrong total number of fruits

Incorrect: Treating this as selecting 2 fruits from a combined 14-fruit collection
This ignores the constraint that we must select exactly one fruit from each basket. The baskets are separate, and we're making independent selections from each.

The Pattern Behind This

This problem demonstrates a fundamental principle in probability: when an event can occur in multiple mutually exclusive ways, we add the probabilities of each way.

For baskets with (a₁ apples, o₁ oranges) and (a₂ apples, o₂ oranges):

P(one apple, one orange) = (a₁/(a₁+o₁)) × (o₂/(a₂+o₂)) + (o₁/(a₁+o₁)) × (a₂/(a₂+o₂))

Notice the elegant symmetry: the probability equals the product of "opposite" fractions from each basket. This pattern holds for any two-basket selection problem where we want exactly one item of each type.

Important limitation: This formula only works when selecting exactly one item from each of exactly two groups. For three or more groups, or for selecting multiple items from the same group, the calculation becomes more complex.

How to Spot This Problem Type

Look for these key phrases that signal an independent selections probability problem:

  • "Select one [item] from each [group]" — indicates independent selections
  • "One of each type" — signals case analysis is needed
  • "Different baskets/containers/groups" — confirms selections are independent
  • "What is the probability that one will be X and the other will be Y" — classic wording for this setup

The structure is always: multiple containers, one selection from each, asking for probability of getting different types. This appears frequently in standardized tests and introductory probability courses.

What If?

1
Change the Numbers
Basket 1 contains 6 apples and 3 oranges, while Basket 2 contains 4 apples and 8 oranges. If you select one fruit from each basket, what is the probability of getting one apple and one orange?
Step 1 — Identify the cases

Case A: Apple from Basket 1, Orange from Basket 2
Case B: Orange from Basket 1, Apple from Basket 2

Step 2 — Calculate Case A

P(Apple from B1) = 6/9 = 2/3
P(Orange from B2) = 8/12 = 2/3
P(Case A) = (2/3) × (2/3) = 4/9

Step 3 — Calculate Case B

P(Orange from B1) = 3/9 = 1/3
P(Apple from B2) = 4/12 = 1/3
P(Case B) = (1/3) × (1/3) = 1/9

Step 4 — Add the cases

P(one apple and one orange) = 4/9 + 1/9 = 5/9

Verification

Check: Both apples: (2/3)(1/3) = 2/9, One each: 5/9, Both oranges: (1/3)(2/3) = 2/9
Total: 2/9 + 5/9 + 2/9 = 9/9 = 1 ✓

Answer: 5/9 ≈ 0.556 or about 55.6%

2
Add a Third Basket
You have three baskets: B1 (4 apples, 2 oranges), B2 (3 apples, 5 oranges), B3 (2 apples, 4 oranges). Selecting one fruit from each basket, what's the probability of getting exactly 2 apples and 1 orange?
Step 1 — Identify successful outcomes

We need exactly 2 apples and 1 orange. This can happen three ways:
AAO, AOA, or OAA

Step 2 — Calculate AAO (Apple, Apple, Orange)

P(A from B1) × P(A from B2) × P(O from B3)
= (4/6) × (3/8) × (4/6) = 48/288 = 1/6

Step 3 — Calculate AOA (Apple, Orange, Apple)

P(A from B1) × P(O from B2) × P(A from B3)
= (4/6) × (5/8) × (2/6) = 40/288 = 5/36

Step 4 — Calculate OAA (Orange, Apple, Apple)

P(O from B1) × P(A from B2) × P(A from B3)
= (2/6) × (3/8) × (2/6) = 12/288 = 1/24

Step 5 — Add all cases

Convert to common denominator 72:
1/6 + 5/36 + 1/24 = 12/72 + 10/72 + 3/72 = 25/72

Answer: 25/72 ≈ 0.347 or about 34.7%

3
Reverse the Problem
Basket 1 has 4 apples and some number of oranges. Basket 2 has 3 apples and 5 oranges. If the probability of selecting one apple and one orange is exactly 1/2, how many oranges are in Basket 1?
Step 1 — Set up the equation

Let x = number of oranges in Basket 1
Basket 1 total fruits: 4 + x
We need: P(one apple, one orange) = 1/2

Step 2 — Write the probability expression

Case A: Apple from B1, Orange from B2
P(Case A) = (4/(4+x)) × (5/8)

Case B: Orange from B1, Apple from B2
P(Case B) = (x/(4+x)) × (3/8)

Step 3 — Set up the equation

(4/(4+x)) × (5/8) + (x/(4+x)) × (3/8) = 1/2

Factor out common terms:
(1/(4+x)) × (20/8 + 3x/8) = 1/2
(20 + 3x)/(8(4+x)) = 1/2

Step 4 — Solve for x

Cross multiply:
2(20 + 3x) = 8(4 + x)
40 + 6x = 32 + 8x
8 = 2x
x = 4

Verification

With 4 oranges in Basket 1:
P = (4/8) × (5/8) + (4/8) × (3/8) = 20/64 + 12/64 = 32/64 = 1/2 ✓

Answer: 4 oranges in Basket 1

4
Without Replacement Twist
From Basket 1 (4 apples, 2 oranges), select 2 fruits without replacement, then select 1 fruit from Basket 2 (3 apples, 5 oranges). What's the probability of ending up with exactly 1 apple and 1 orange total?
Step 1 — Identify the successful scenarios

We need exactly 1 apple and 1 orange total from 3 fruits.
Case A: 2 apples from B1, then 1 orange from B2
Case B: 1 apple + 1 orange from B1, then 1 apple from B2
Case C: 1 apple + 1 orange from B1, then 1 orange from B2
Case D: 2 oranges from B1, then 1 apple from B2

Step 2 — Calculate each case

Cases A and D fail (give us 2 of same type).

Case B: AA from B1, then O from B2
P = (4/6) × (3/5) × (5/8) = 60/240 = 1/4

Step 3 — Actually, reconsider the constraint

Wait - we want exactly 1 apple and 1 orange total.
This means selecting 2 fruits from B1 already gives us too many fruits if we want just 1 of each type total.

Step 4 — Correct interpretation

For exactly 1 apple and 1 orange total (2 fruits), we need:
- 1 fruit from B1 + 1 fruit from B2, OR
- 2 fruits from B1 + 0 from B2, OR
- 0 fruits from B1 + 2 from B2

Since we must select 2 from B1 and 1 from B2, this is impossible.

Reinterpret as: exactly 1 apple among the 3 fruits

Case A: AA from B1, O from B2 → 2A, 1O (fails)
Case B: AO from B1, A from B2 → 2A, 1O (fails)
Case C: AO from B1, O from B2 → 1A, 2O (fails)
Case D: OO from B1, A from B2 → 1A, 2O (fails)

Answer: 0 (impossible with the given constraints)

Frequently Asked Questions

+ How do you find the probability of getting one of each type when selecting from different groups?
Use case analysis to break down all successful outcomes. In this example, success means (Apple from Basket 1, Orange from Basket 2) OR (Orange from Basket 1, Apple from Basket 2). Calculate each case separately: (4/6) × (5/8) + (2/6) × (3/8) = 20/48 + 6/48 = 26/48 = 13/24.
+ When do you add probabilities versus multiply them?
Multiply probabilities for AND events (things that must both happen). Add probabilities for OR events (mutually exclusive alternatives). Here we multiply within each case (apple AND orange) but add between cases (Case A OR Case B).
+ What's the difference between independent events and mutually exclusive events?
Independent events don't affect each other's probability (selecting from separate baskets). Mutually exclusive events can't happen at the same time (getting apple from Basket 1 OR orange from Basket 1). This problem involves both concepts.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-10-28

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