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# drawing straws

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In each of three boxes are four straws with integer lengths  cm,  cm,  cm, and  cm. A straw is randomly selected from each box. What is the probability that a triangle can be formed with the three straws chosen?

Jul 13, 2023

#1
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The sum of any two sides of a triangle must be greater than its third side.

That's the explanation.  The length of the straws wasn't visible in the post.

Jul 13, 2023
#2
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ummm oops changing that now

moneydude242  Jul 13, 2023
#3
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OK corrected version:

In each of three boxes are four straws with integer lengths 1 cm, 2 cm, 3 cm, and 4 cm. A straw is randomly selected from each box. What is the probability that a triangle can be formed with the three straws chosen?

Jul 13, 2023
#4
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Here is an explanation of the answer:

We can't form a triangle if any of the straws have length 1, so we need to consider the cases where all three straws have length 2, 2 and 3, or 3 and 4.

Case 1: All three straws have length 2. There are (34​)=4 ways to choose which straws to select, so the probability of this case is 4/4^3=1/64

Case 2: Two straws have length 2 and one has length 3. There are (24​)=6 ways to choose which straws to select, so the probability of this case is 6/4^3​=3/64​.

Case 3: One straw has length 2 and the other two have length 3. There are (14​)=4 ways to choose which straw has length 2, so the probability of this case is 4/4^3=1/64​​.

The total probability is then 1/64​+3/64​+1/64​=5/64​​.

Guest Jul 13, 2023
#5
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I count 13 valid triangles as follows:

(1, 1, 1), (1, 2, 2), (1, 3, 3), (1, 4, 4), (2, 2, 2), (2, 2, 3), (2, 3, 3), (2, 3, 4), (2, 4, 4), (3, 3, 3), (3, 3, 4), (3, 4, 4), (4, 4, 4)= 13 triangles possible

Total of all triangles =4^3 =64

Therefore the probability is: 13 / 64

Jul 15, 2023