Probability Math Example 4

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

Example 4

medium
A box contains 44 red, 55 blue, and 11 green marble. Two marbles are drawn without replacement. What is the probability of getting one red and one blue in any order?

Solution

  1. 1
    Count the two valid orders: red then blue, or blue then red.
  2. 2
    P(R then B)=410⋅59P(R\text{ then }B) = \frac{4}{10} \cdot \frac{5}{9} and P(B then R)=510⋅49P(B\text{ then }R) = \frac{5}{10} \cdot \frac{4}{9}.
  3. 3
    Add them: 410⋅59+510⋅49=2090+2090=4090=49\frac{4}{10} \cdot \frac{5}{9} + \frac{5}{10} \cdot \frac{4}{9} = \frac{20}{90} + \frac{20}{90} = \frac{40}{90} = \frac{4}{9}.

Answer

P(one red and one blue)=49P(\text{one red and one blue}) = \frac{4}{9}
When two different orders produce the same event, add the probabilities of each order. Without replacement, the second probability changes because the total number of marbles decreases.

About Probability

Probability is a number between 0 and 1 (inclusive) that measures how likely an event is to occur, where 0 means impossible and 1 means certain.

Learn more about Probability →

More Probability Examples

Example 1 easy

A bag contains [formula] red, [formula] blue, and [formula] green marbles. What is the probability o

Example 2 medium

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Example 3 medium

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Example 5 medium

A bag contains 4 red, 3 blue, and 5 green marbles. What is the probability of drawing a red or blue

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