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question:A board game spinner is divided into three parts labeled A, B and C. The probability of the spinner landing on A is frac{1}{3} and the probability of the spinner landing on B is frac{5}{12}. What is the probability of the spinner landing on C? Express your answer as a common fraction.
answer:To find the probability of the spinner landing on C, I need to subtract the probabilities of the spinner landing on A and B from 1, since the sum of the probabilities of all possible outcomes is 1. I can write this as an equation: P(C) = 1 - P(A) - P(B). I know that P(A) = frac{1}{3} and P(B) = frac{5}{12}, so I can plug those values into the equation and simplify. I get: P(C) = 1 - frac{1}{3} - frac{5}{12} = frac{12}{12} - frac{4}{12} - frac{5}{12} = frac{3}{12}. I can reduce this fraction by dividing the numerator and denominator by 3, and I get: P(C) = frac{1}{4}.
question:My school's math club has 6 boys and 8 girls. I need to select a team to send to the state math competition. We want 6 people on the team. In how many ways can I select the team without restrictions?
answer:I need to choose 6 people out of 14, and the order does not matter. This is a combination problem, not a permutation problem. The formula for combinations is nCr = n! / (r! * (n-r)!), where n is the total number of choices and r is the number of selections. Plugging in the numbers, I get 14C6 = 14! / (6! * 8!) = 3003.
question:w many 4letter words with at least one consonant can be constructed from the letters A, B, C, D, and E? (Note that B, C, and D are consonants, any word is valid, not just English language words, and letters may be used more than once.)
answer:First we count the number of all 4-letter words with no restrictions on the word. Then we count the number of 4-letter words with no consonants. We then subtract to get the answer. Each letter of a word must be one of A, B, C, D, or E, so the number of 4-letter words with no restrictions on the word is 5times 5times 5times 5=625. Each letter of a word with no consonant must be one of A or E. So the number of all 4-letter words with no consonants is 2times 2times 2times 2=16. Therefore, the number of 4-letter words with at least one consonant is 625-16=609.
question:Melinda will roll two standard sixsided dice and make a twodigit number with the two numbers she rolls. For example, if she rolls a 6 and a 3, she can either form 36 or 63. What is the probability that she will be able to make an integer between 10 and 20, inclusive? Express your answer as a common fraction.
answer:She can do this if and only if at least one of the dice lands on a 1. The probability neither of the dice is a 1 is left(frac{5}{6}right) left(frac{5}{6}right) = frac{25}{36}. So the probability at least one die is a 1 is 1-frac{25}{36} = frac{11}{36}.