Question

a single die is rolled twice. find the probability of rolling an odd number and a number greater than 3 in either order. the probability of rolling an odd number and a number greater than 3 is (type an integer or a simplified fraction)

Explanation:

Step1: Determine sample - space size

When a die is rolled twice, the sample - space size $n(S)=6\times6 = 36$ since there are 6 possible outcomes on the first roll and 6 possible outcomes on the second roll.

Step2: Define the events

Let $A$ be the set of odd numbers on a die $\{1,3,5\}$ and $B$ be the set of numbers greater than 3 on a die $\{4,5,6\}$.

Step3: Calculate the number of favorable outcomes

Case 1: First roll is odd and second roll is greater than 3.
The number of odd numbers is 3, and the number of numbers greater than 3 is 3. The number of outcomes for (odd, > 3) is $3\times3=9$ (e.g., (1,4), (1,5), (1,6), (3,4), (3,5), (3,6), (5,4), (5,5), (5,6)).
Case 2: First roll is greater than 3 and second roll is odd.
The number of outcomes for (> 3, odd) is $3\times3 = 9$ (e.g., (4,1), (4,3), (4,5), (5,1), (5,3), (5,5), (6,1), (6,3), (6,5)). But we have double - counted the outcomes where the number is both odd and greater than 3 (i.e., (5,5)) once in each case. The number of such double - counted outcomes is 3 ( (5,5), (5,5) is counted twice; in general, the common elements of $A$ and $B$ are 5, and the pairs (5,5) is counted in both cases).
The number of favorable outcomes $n(E)=9 + 9-3=15$.

Step4: Calculate the probability

The probability $P(E)=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}$.

Answer:

$\frac{5}{12}$