A single die is rolled twice. Find the probability of rolling an even number the first time and a number greater than 1 the second time.

1 answer

To find the probability of rolling an even number on the first roll and a number greater than 1 on the second roll with a single die, let's break down the problem:

  1. Determine the outcomes for each roll:

    • A standard die has six faces showing the numbers 1, 2, 3, 4, 5, and 6.
  2. Calculate the probability of rolling an even number on the first roll:

    • The even numbers on a die are 2, 4, and 6.
    • There are 3 even numbers out of a total of 6 possible outcomes.

    \[ P(\text{even on first roll}) = \frac{3}{6} = \frac{1}{2} \]

  3. Calculate the probability of rolling a number greater than 1 on the second roll:

    • The numbers greater than 1 on a die are 2, 3, 4, 5, and 6.
    • There are 5 numbers greater than 1 out of a total of 6 possible outcomes.

    \[ P(\text{number > 1 on second roll}) = \frac{5}{6} \]

  4. Calculate the combined probability:

    • Since the two rolls are independent events, the total probability can be found by multiplying the probabilities of each event.

    \[ P(\text{even on first roll and number > 1 on second roll}) = P(\text{even on first roll}) \times P(\text{number > 1 on second roll}) = \frac{1}{2} \times \frac{5}{6} \]

    \[ = \frac{5}{12} \]

Thus, the probability of rolling an even number the first time and a number greater than 1 the second time is

\[ \boxed{\frac{5}{12}}. \]