| [ | |
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| "input": "", | |
| "output": "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: $P(C) = \\frac{1}{4}$. ", | |
| "instruction": "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." | |
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| "output": "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. ", | |
| "instruction": "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?" | |
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| "output": "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.\n\nEach 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 $5\\times 5\\times 5\\times 5=625$. 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 $2\\times 2\\times 2\\times 2=16$. Therefore, the number of 4-letter words with at least one consonant is $625-16=609$.", | |
| "instruction": "How many 4-letter 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.)" | |
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| "output": "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 $1-\\frac{25}{36} = \\frac{11}{36}$.", | |
| "instruction": "Melinda will roll two standard six-sided dice and make a two-digit 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." | |
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| "input": "", | |
| "output": "Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the sequence is blocks of $1$ to $4$ H's separated by T's and ending in $5$ H's. Since the first letter could be T or the sequence could start with a block of H's, the total probability is that $3/2$ of it has to start with an H.\nThe answer to the problem is then the sum of all numbers of the form $\\frac 32 \\left( \\frac 1{2^a} \\cdot \\frac 12 \\cdot \\frac 1{2^b} \\cdot \\frac 12 \\cdot \\frac 1{2^c} \\cdots \\right) \\cdot \\left(\\frac 12\\right)^5$, where $a,b,c \\ldots$ are all numbers $1-4$, since the blocks of H's can range from $1-4$ in length. The sum of all numbers of the form $(1/2)^a$ is $1/2+1/4+1/8+1/16=15/16$, so if there are n blocks of H's before the final five H's, the answer can be rewritten as the sum of all numbers of the form $\\frac 32\\left( \\left(\\frac {15}{16}\\right)^n \\cdot \\left(\\frac 12\\right)^n \\right) \\cdot \\left(\\frac 1{32}\\right)=\\frac 3{64}\\left(\\frac{15}{32}\\right)^n$, where $n$ ranges from $0$ to $\\infty$, since that's how many blocks of H's there can be before the final five. This is an infinite geometric series whose sum is $\\frac{3/64}{1-(15/32)}=\\frac{3}{34}$, so the answer is $37$.", | |
| "instruction": "Let $p$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p$ can be written in the form $m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.\n" | |
| } | |
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