Question

charles has two number cubes. the faces of each number cube are numbered from 1 to 6. charles will roll each number cube one time. move the correct answer to each box to show the probability of rolling a 5 on the first number cube and an even number on the second number cube. not all answers will be used.

Explanation:

Answer:

To solve the probability of rolling a 5 on the first number cube and an even number on the second number cube, we follow these steps:

Step 1: Determine the probability of rolling a 5 on the first cube

A standard number cube (die) has 6 faces numbered from 1 to 6. There is only 1 face with the number 5. So the probability of rolling a 5 is:
\( P(5) = \frac{1}{6} \)

Step 2: Determine the probability of rolling an even number on the second cube

The even numbers on a number cube are 2, 4, and 6. So there are 3 favorable outcomes out of 6 total outcomes. The probability of rolling an even number is:
\( P(\text{even}) = \frac{3}{6} = \frac{1}{2} \)

Step 3: Multiply the two probabilities (since the rolls are independent events)

To find the probability of both events happening (rolling a 5 on the first cube AND an even number on the second cube), we multiply the probabilities of each event:
\( P(5 \text{ and even}) = P(5) \times P(\text{even}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \)

So the numerator of the probability fraction is 1 (from the first cube's successful outcome count), the denominator of the first fraction (for the first cube) is 6, the numerator of the second fraction (for the second cube) is 3, the denominator of the second fraction is 6, and the final numerator is 1, final denominator is 12. If we are filling in the boxes (assuming the first box is numerator of first probability, second box denominator of first, third box numerator of second, fourth box denominator of second, and then the result's numerator and denominator), we would have:

First box (probability of 5 numerator): 1

Second box (probability of 5 denominator): 6

Third box (probability of even numerator): 3

Fourth box (probability of even denominator): 6

Final numerator (result): 1

Final denominator (result): 12

If we consider the multiplication of the number of favorable outcomes for each event over the total number of possible outcomes for both events:

Total number of possible outcomes when rolling two cubes: \( 6 \times 6 = 36 \)

Number of favorable outcomes (5 on first and even on second): There is 1 way to get 5 on the first cube, and 3 ways to get an even number on the second cube, so \( 1 \times 3 = 3 \)

Wait, no, earlier calculation was wrong. Wait, no:

Wait, probability of 5 on first cube: 1/6 (1 favorable out of 6)

Probability of even on second cube: 3/6 = 1/2 (3 favorable out of 6)

Multiplying these: (1/6) (3/6) = 3/36 = 1/12. Ah, right, so the number of favorable outcomes is 1 (for 5) 3 (for even) = 3, and total outcomes is 6*6=36. Then 3/36 simplifies to 1/12.

So if the problem is structured as (number of favorable for first event) / (total for first event) * (number of favorable for second event) / (total for second event) = (result numerator) / (result denominator)

Then:

First numerator: 1

First denominator: 6

Second numerator: 3

Second denominator: 6

Result numerator: 3 (or 1 if simplified)

Result denominator: 36 (or 12 if simplified)

Looking at the answer choices (1,2,3,5,6,12,15,30), the possible numbers to fill in would be:

First box (maybe numerator of first probability): 1

Second box (denominator of first probability): 6

Third box (numerator of second probability): 3

Fourth box (denominator of second probability): 6

Then the result would be 3/36 = 1/12, so numerator 1, denominator 12.

So if we need to move the correct numbers, the numbers involved are 1, 6, 3, 6, 1, 12. From the given options (1,2,3,5,6,12,15,30), the relevant ones are 1, 3, 6, 12.

So for example, if the first fraction is (1)/(6), the second is (3)/(6), and the result is (1)/(12) (since 13=3 and 66=36, then 3/36=1/12).

So the numbers to move would be 1 (first numerator), 6 (first denominator), 3 (second numerator), 6 (second denominator), and then 1 (result numerator) and 12 (result denominator).