Conditional Marble Probability

P(B|A) with marble bag examples

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P(R,R,B | At Least 1 Red)

Question

A bag contains 2 red and 2 blue marbles. You draw 3 marbles without replacement. Given that at least one marble is red, what is the probability the sequence was Red, Red, Blue?

Step-by-Step Solution

List All Possible Sequences (Draw 3 from 4)

With 2R and 2B, possible sequences of 3 draws:

RRB

2R, 1B

RBR

2R, 1B

BRR

2R, 1B

RBB

1R, 2B

BRB

1R, 2B

BBR

1R, 2B

Note: BBB is impossible (only 2 blue marbles available). All sequences have at least 1 red.

Calculate P(RRB)

Probability of drawing Red, then Red, then Blue:

1st draw: 2R/4 total

2nd draw: 1R/3 remaining

3rd draw: 2B/2 remaining

Calculate P(At Least 1 Red)

Since BBB is impossible with only 2 blue marbles:

Every valid 3-draw sequence contains at least one red.

Apply Conditional Probability Formula

Answer

Key Insight

In this case, the condition "at least 1 red" is always satisfied (probability = 1), so the conditional probability equals the unconditional probability.

If we had more blue marbles (e.g., 5 blue), then BBB would be possible, and the condition would actually reduce the sample space, changing the answer.