Question

a die is rolled twice. what is the probability of showing a 1 on the first roll and an even number on the second roll? your answer is :

Explanation:

Step1: Probability of getting 1 on first roll

A die has 6 faces. The probability of getting a 1 on the first roll, $P(1)$, is $\frac{1}{6}$ since there is 1 favorable outcome (rolling a 1) out of 6 possible outcomes.

Step2: Probability of getting an even number on second roll

The even numbers on a die are 2, 4, 6. So there are 3 favorable outcomes out of 6 possible outcomes. The probability of getting an even - number on the second roll, $P(E)$, is $\frac{3}{6}=\frac{1}{2}$.

Step3: Use the multiplication rule for independent events

Since the two rolls of the die are independent events, the probability of both events occurring is the product of their individual probabilities. Let $P$ be the probability of showing a 1 on the first roll and an even number on the second roll. Then $P = P(1)\times P(E)$.
Substitute $P(1)=\frac{1}{6}$ and $P(E)=\frac{1}{2}$ into the formula: $P=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}$.

Answer:

$\frac{1}{12}$