Let C be the number of cats and D be the number of dogs. We are given that \(D = C + 8\) and \(D + C = 30\). We can substitute the first equation into the second: \((C + 8) + C = 30\). This simplifies to \(2C + 8 = 30\). Subtract 8 from both sides: \(2C = 22\). Divide by 2: \(C = 11\). There are 11 cats. The total number of animals is 30. The probability of choosing a cat is the number of cats divided by the total number of animals: \(\frac{11}{30}\).

The possible outcomes when rolling a six-sided die are {1, 2, 3, 4, 5, 6}. We need to find the outcomes that satisfy two conditions: being an even number and being a factor of 12. The even numbers are {2, 4, 6}. Now we check which of these are also factors of 12. 12 is divisible by 2, 4, and 6. So, all three numbers {2, 4, 6} satisfy both conditions. There are 3 favorable outcomes. The total number of outcomes is 6. The probability is \(\frac{3}{6}\), which simplifies to \(\frac{1}{2}\).

There are two ways to solve this. Method 1: Find the number of crayons that are not blue. The total is 50, and 10 are blue, so \(50 - 10 = 40\) crayons are not blue. The probability is \(\frac{40}{50}\), which simplifies to \(\frac{4}{5}\). Method 2: Find the probability of picking a blue crayon, which is \(\frac{10}{50} = \frac{1}{5}\). The probability of the event NOT happening is 1 minus the probability of the event happening. So, the probability of not picking a blue crayon is \(1 - \frac{1}{5} = \frac{4}{5}\).

First, find the total number of slips for weekend days (Saturday and Sunday). This is \(10 + 6 = 16\). Next, find the number of slips for weekdays by subtracting the weekend slips from the total: \(30 - 16 = 14\). The probability of choosing a weekday is the number of weekday slips divided by the total number of slips, which is \(\frac{14}{30}\). This fraction can be simplified by dividing both the numerator and the denominator by 2: \(\frac{14 \div 2}{30 \div 2} = \frac{7}{15}\).

First, calculate the number of chocolate chip cookies: \(\frac{1}{3} \times 36 = 12\). Next, calculate the number of oatmeal cookies: \(\frac{1}{4} \times 36 = 9\). The total number of chocolate chip and oatmeal cookies is \(12 + 9 = 21\). To find the number of peanut butter cookies, subtract this from the total: \(36 - 21 = 15\). The probability of choosing a peanut butter cookie is the number of peanut butter cookies divided by the total number of cookies: \(\frac{15}{36}\). To simplify, divide the numerator and denominator by their greatest common factor, which is 3: \(\frac{15 \div 3}{36 \div 3} = \frac{5}{12}\).

First, find the number of green marbles. There are 24 marbles in total, with 8 red and 6 blue. The number of red and blue marbles together is \(8 + 6 = 14\). The number of green marbles is the total minus the red and blue marbles: \(24 - 14 = 10\) green marbles. The probability of choosing a green marble is the number of green marbles divided by the total number of marbles, which is \(\frac{10}{24}\). This fraction must be simplified. Both 10 and 24 are divisible by 2, so \(\frac{10}{24} = \frac{5}{12}\).

The question asks for the probability that a randomly selected student has brown hair. We need to find the total number of students with brown hair and divide it by the total number of students. There are 5 boys with brown hair and 8 girls with brown hair. So, the total number of students with brown hair is \(5 + 8 = 13\). The total number of students in the class is 25. The probability is the number of students with brown hair divided by the total number of students: \(\frac{13}{25}\). The information that there are 12 boys is extra information not needed to find the total number of students with brown hair.

Let Y be the number of yellow gumballs and R be the number of red gumballs. We know that \(R = 2Y\) and \(R + Y = 36\). Substitute the first equation into the second: \(2Y + Y = 36\), which simplifies to \(3Y = 36\). Dividing by 3, we find \(Y = 12\). So there are 12 yellow gumballs. The probability of getting a yellow gumball is the number of yellow gumballs divided by the total number of gumballs: \(\frac{12}{36}\). This fraction simplifies to \(\frac{1}{3}\).

First, identify the total number of possible outcomes, which is 20 since there are 20 cards. Next, find the number of favorable outcomes. The multiples of 4 between 1 and 20 are 4, 8, 12, 16, and 20. There are 5 such multiples. The multiples of 7 between 1 and 20 are 7 and 14. There are 2 such multiples. Since there are no numbers that are multiples of both 4 and 7 in this range, we can add the number of favorable outcomes together: \(5 + 2 = 7\). The probability is the number of favorable outcomes divided by the total number of outcomes, which is \(\frac{7}{20}\).

First, find the total number of sections on the spinner. We are told the probability of landing on red is \(\frac{1}{4}\) and that there are 5 red sections. This means that 5 is \(\frac{1}{4}\) of the total number of sections. So, the total number of sections is \(5 \times 4 = 20\). Next, find the number of blue sections. There are 2 more blue sections than red sections, so the number of blue sections is \(5 + 2 = 7\). The probability of landing on blue is the number of blue sections divided by the total number of sections, which is \(\frac{7}{20}\).