EZ Math

Statistics · Probability

How to Find the Probability of Two Independent Events

Problem

heads, then a 6 on one die\text{heads, then a 6 on one die}

Answer

112\frac{1}{12}

For independent events, find each probability, then multiply. The events are independent when one result does not change the chance of the other.

Step-by-step solution

  1. Write the probability of each event.

    P(heads)=12,P(6)=16P(\text{heads}) = \frac{1}{2},\quad P(6) = \frac{1}{6}

    Why: The coin and the die do not affect each other.

  2. Multiply the two probabilities.

    12×16=112\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

    Why: For independent events, the probabilities multiply.

  3. Write the answer.

    P(heads then 6)=112P(\text{heads then } 6) = \frac{1}{12}

    Why: One of twelve equally likely pairs gives this result.

Common mistakes

  • Adding the probabilities instead of multiplying.
  • Multiplying the counts of outcomes wrongly. Multiply the two fractions.
  • Treating the events as dependent. A coin toss does not change the die.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

  1. two heads in two tosses\text{two heads in two tosses}

    Show answer

    14\frac{1}{4}

  2. heads, then a 3 on one die\text{heads, then a 3 on one die}

    Show answer

    112\frac{1}{12}

  3. a 5, then a 6 on two dice\text{a 5, then a 6 on two dice}

    Show answer

    136\frac{1}{36}

  4. two even numbers on two dice\text{two even numbers on two dice}

    Show answer

    14\frac{1}{4}

  5. a blue, then a red, with 13 and 12\text{a blue, then a red, with } \frac{1}{3} \text{ and } \frac{1}{2}

    Show answer

    16\frac{1}{6}