Compound Events Statistics Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

medium
A bag contains 4 red and 6 blue marbles. You draw two marbles without replacement. What is the probability that both are red?

Solution

  1. 1
    Step 1: Without replacement, the draws are dependent — the first draw changes the composition.
  2. 2
    Step 2: P(1st red)=410=25P(\text{1st red}) = \frac{4}{10} = \frac{2}{5}.
  3. 3
    Step 3: After drawing a red marble, 3 red and 6 blue remain (9 total). P(2nd red∣1st red)=39=13P(\text{2nd red} | \text{1st red}) = \frac{3}{9} = \frac{1}{3}. So P(both red)=25×13=215P(\text{both red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}.

Answer

P(both red)=215≈0.133P(\text{both red}) = \frac{2}{15} \approx 0.133.
Without replacement, events are dependent because the outcome of the first event changes the probabilities for the second. The multiplication rule for dependent events uses conditional probability: P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B|A).

About Compound Events

Compound events are probability events made up of two or more simple events combined using 'and' (both events occur) or 'or' (at least one occurs). For independent 'and' events, multiply probabilities; for 'or' events, add probabilities and subtract any overlap.

Learn more about Compound Events →

More Compound Events Examples

Example 1 easy

A bag contains 3 red and 5 blue marbles. You draw one marble, replace it, then draw another. What is

Example 3 medium

A spinner has sections: Red (50%), Blue (30%), Green (20%). You spin twice. What is the probability

Example 4 hard

A deck has 52 cards. You draw 2 cards without replacement. What is the probability of drawing at lea

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