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### 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 $m+n$. ### Response:

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. The 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: I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different marbles at random. What is the expected value of the sum of the numbers on the marbles? ### Response:

I need to find the average of all possible sums of two different marbles from the bag. To do that, I can list all the possible outcomes and their probabilities, and then multiply each outcome by its probability and add them up. There are 5 choose 2, or 10, ways to take out two different marbles from the bag. Each way has the same probability of 1/10, since I'm assuming the bag is well-mixed and each marble has an equal chance of being picked. The possible outcomes and their sums are: (1, 2) with sum 3, (1, 3) with sum 4, (1, 4) with sum 5, (1, 5) with sum 6, (2, 3) with sum 5, (2, 4) with sum 6, (2, 5) with sum 7, (3, 4) with sum 7, (3, 5) with sum 8, and (4, 5) with sum 9. So the expected value of the sum is (3 + 4 + 5 + 6 + 5 + 6 + 7 + 7 + 8 + 9) / 10, which is 60 / 10, or 6. This makes sense, because the average of the numbers on the marbles is 3, and I expect the average of the sum of two marbles to be twice that.

### Instruction: The number of increasing sequences of positive integers $a_1 \le a_2 \le a_3 \le \cdots \le a_{10} \le 2007$ such that $a_i-i$ is even for $1\le i \le 10$ can be expressed as ${m \choose n}$ for some positive integers $m > n$. Compute the remainder when $m$ is divided by 1000. ### Response:

The numbers $a_i - i$ are ten not-necessarily distinct even elements of the set $\{0, 1, 2, \ldots, 1997\}$. Moreover, given ten not-necessarily distinct elements of $\{0, 1, 2, \ldots, 1997\}$, we can reconstruct the list $a_1, a_2, \ldots, a_{10}$ in exactly one way, by adding 1 to the smallest, then adding 2 to the second-smallest (which might actually be equal to the smallest), and so on. Thus, the answer is the same as the number of ways to choose 10 elements with replacement from the set $\{0, 2, 4, \ldots, 1996\}$, which has 999 elements. This is a classic problem of combinatorics; in general, there are ${m + n - 1 \choose m}$ ways to choose $m$ things from a set of $n$ with replacement. In our case, this gives the value of ${999 + 10 - 1 \choose 10} = {1008 \choose 10}$, so the answer is $8$.

### 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. ### Response:

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: Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. ### Response:

There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting). Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$. Therefore, there are $999 - (99 + 162) = 738$ such ordered pairs.

### 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.) ### Response:

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 $5\times 5\times 5\times 5=625$. 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 four-digit numbers greater than 2999 can be formed such that the product of the middle two digits exceeds 5? ### Response:

For the first digit, there are seven choices (3, 4, 5, 6, 7, 8, or 9). For the last digit, there are ten choices (0 through 9). We know that if either of the middle digits is 0, their product will not exceed 5. So, only consider pairs of middle digits formed from choosing two numbers between 1 and 9, inclusive. There are $9 \cdot 9$ such pairs possible. The only pairs whose product will not exceed 5 are 11, 12, 13, 14, 15, 21, 22, 31, 41, and 51. Thus, there are $9 \cdot 9 - 10 = 71$ ways in which we can choose the middle two digits. Thus, there are $ 7 \cdot 71 \cdot 10 = 4970$ such numbers.

### 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. ### Response:

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: 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? ### Response:

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: A box contains 5 white balls and 6 black balls. Two balls are drawn out of the box at random. What is the probability that they both are white? ### Response:

There are $\binom{11}{2} = 55$ combinations of two balls that can be drawn. There are $\binom{5}{2} = 10$ combinations of two white balls that can be drawn. So the probability that two balls pulled out are both white is $\dfrac{10}{55} = \dfrac{2}{11}$.