Question
A bag contains 3 red tokens and 2 blue tokens. Two tokens are drawn without replacement. What is the probability that both are red?
- a.3/10
- b.9/25
- c.3/5
- d.1/2
Correct answer
A. 3/10
Full reasoning
- 1The first draw is red with probability 3/5.
- 2Given a red first draw, the bag now contains 2 red tokens among 4 total tokens.
- 3Multiply along this path: (3/5)(2/4) = 3/10.
- 4As a separate check, 3 of the 10 unordered token pairs contain two red tokens. If the first token were replaced, 9/25 would instead be correct.
Why each choice is right or wrong
Choice A
Correct. A red first draw has probability 3/5; conditional on that result, the second red draw has probability 2/4.
Choice B
This reuses 3/5 for the second draw as if the first token had been replaced.
Choice C
This accounts only for a red first draw. The question requires both draws to be red, so a second conditional factor is needed.
Choice D
This is the probability of a red second draw conditional on an already-red first draw, not the joint event.
Related formula
Multiplication rule
P(A and B) = P(A) P(B | A)
Assumptions: The conditional probability is defined because P(A) is positive. Draws without replacement are not independent.
Related topic and practice
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Sources
Original authored exercise; no database question ID and no protected or official exam-bank content. Specialist review pending.
- Introductory Statistics 2e, Section 3.3: Two Basic Rules of Probability — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY 4.0; attribute and avoid verbatim reuse beyond short cited references.
Sources and corrections
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