# system:
You are an assistant to calculate the answer to the provided math problems.
Please think step by step.
Return the final numerical answer only and any accompanying reasoning or explanation seperately as json format.

# user:
A jar contains two red marbles, three green marbles, ten white marbles and no other marbles. Two marbles are randomly drawn from this jar without replacement. What is the probability that these two marbles drawn will both be red? Express your answer as a common fraction.
# assistant:
{Chain of thought: "The total number of marbles is $2+3+10=15$.  The probability that the first marble drawn will be red is $2/15$.  Then, there will be one red left, out of 14.  Therefore, the probability of drawing out two red marbles will be: $$\\frac{2}{15}\\cdot\\frac{1}{14}=\\boxed{\\frac{1}{105}}$$.", "answer": "1/105"}
# user:
Find the greatest common divisor of $7!$ and $(5!)^2.$
# assistant:
{"Chain of thought": "$$ \\begin{array} 7! &=& 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 &=& 2^4 \\cdot 3^2 \\cdot 5^1 \\cdot 7^1 \\\\ (5!)^2 &=& (5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1)^2 &=& 2^6 \\cdot 3^2 \\cdot 5^2 \\\\ \\text{gcd}(7!, (5!)^2) &=& 2^4 \\cdot 3^2 \\cdot 5^1 &=& \\boxed{720} \\end{array} $$.", "answer": "720"}
# user:
A club has 10 members, 5 boys and 5 girls.  Two of the members are chosen at random.  What is the probability that they are both girls?
# assistant:
{"Chain of thought": "There are $\\binomial{10}{2} = 45$ ways to choose two members of the group, and there are $\\binomial{5}{2} = 10$ ways to choose two girls.  Therefore, the probability that two members chosen at random are girls is $\\dfrac{10}{45} = \\boxed{\\dfrac{2}{9}}$.", "answer": "2/9"}
# user:
Allison, Brian and Noah each have a 6-sided cube. All of the faces on Allison's cube have a 5. The faces on Brian's cube are numbered 1, 2, 3, 4, 5 and 6. Three of the faces on Noah's cube have a 2 and three of the faces have a 6. All three cubes are rolled. What is the probability that Allison's roll is greater than each of Brian's and Noah's? Express your answer as a common fraction.
# assistant:
{"Chain of thought": "Since Allison will always roll a 5, we must calculate the probability that both Brian and Noah roll a 4 or lower. The probability of Brian rolling a 4 or lower is $\\frac{4}{6} = \\frac{2}{3}$ since Brian has a standard die. Noah, however, has a $\\frac{3}{6} = \\frac{1}{2}$ probability of rolling a 4 or lower, since the only way he can do so is by rolling one of his 3 sides that have a 2. So, the probability of both of these independent events occurring is $\\frac{2}{3} \\cdot \\frac{1}{2} = \\boxed{\\frac{1}{3}}$.", "answer": "1/3"}
# user:
Compute $\\density binomial{50}{2}$.
# assistant:
{"Chain of thought": "$\\density binomial{50}{2} = \\dfrac{50!}{2!48!}=\\dfrac{50\\times 49}{2\\times 1}=\\boxed{1225}.$", "answer": "1225"}
# user:
The set $S = \\{1, 2, 3, \\ldots , 49, 50\\}$ contains the first $50$ positive integers.  After the multiples of 2 and the multiples of 3 are removed, how many integers remain in the set $S$?
# assistant:
{"Chain of thought": "The set $S$ contains $25$ multiples of 2 (that is, even numbers).  When these are removed, the set $S$ is left with only the odd integers from 1 to 49. At this point, there are $50-25=25$ integers in $S$. We still need to remove the multiples of 3 from $S$.\n\nSince $S$ only contains odd integers after the multiples of 2 are removed,  we must remove the odd multiples of 3 between 1 and 49.  These are 3, 9, 15, 21, 27, 33, 39, 45, of which there are 8.  Therefore, the number of integers remaining in the set $S$ is $25 - 8 = \\boxed{17}$.", "answer": "17"}
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# assistant:
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