Basic probability without replacement: A box has 4 white and 3 red balls. Two balls are drawn successively without replacement. What is the probability that both are white?

Introduction / Context:
Sampling without replacement is a fundamental probability model. Here, drawing two balls successively from a small urn illustrates conditional probability and the multiplication rule.

Given Data / Assumptions:

• Total balls: 7 (4 white, 3 red).
• Two draws in succession without replacement.
• Event of interest: both balls are white.

Concept / Approach:

Use the multiplication rule: P(A ∩ B) = P(A) * P(B | A). Let A be “first ball white” and B be “second ball white given first white”.

Step-by-Step Solution:

Verification / Alternative check:

Combinatorial approach: favorable outcomes = C(4,2) = 6; total 2-ball selections = C(7,2) = 21; so probability = 6/21 = 2/7. Matches conditional method.

Why Other Options Are Wrong:

• 4/21 and 8/21 are common arithmetic slips; 3/7 counts single white outcomes incorrectly; 1/7 is too low.

Common Pitfalls:

• Accidentally using replacement or forgetting to update counts on the second draw.
• Confusing ordered draws with unordered combinations.

Final Answer:

2/7