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Home » We want the probability that **at least two** channels allow sufficient flow, i.e., - Londonproperty

We want the probability that at least two channels allow sufficient flow, i.e., - Londonproperty

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 oxed{rac{1}{2}}

📌 Question: A museum curator is aligning two historical astrolabes represented by vectors $ egin{pmatrix} \cos lpha \ \sin lpha \end{pmatrix} $ and $ egin{pmatrix} \cos eta \ \sin eta \end{pmatrix} $ in the plane. If the cosine of the angle between them is $ rac{1}{3} $, and $ lpha + eta = 90^\circ $, find $ \sin(2lpha) $.

📌 The angle between two unit vectors $ \mathbf{u} $ and $ \mathbf{v} $ is given by $ \cos heta = \mathbf{u} \cdot \mathbf{v} $. So,

📌 \cos(lpha - eta) = rac{1}{3}.

📌 We are given $ lpha + eta = 90^\circ $, so $ eta = 90^\circ - lpha $. Substituting,

📌 \cos(lpha - (90^\circ - lpha)) = \cos(2lpha - 90^\circ) = rac{1}{3}.

📌 Using the identity $ \cos(90^\circ - x) = \sin x $, but here we use:

📌 \cos(2lpha - 90^\circ) = \sin(2lpha),

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Question: Three distinct prime numbers less than 50 are selected at ra...
Solution: First, note that the sum of three numbers is even if and onl...
List all prime numbers less than 50:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
There are 15 primes in total, of which 14 are odd and 1 (2) is even.
Total number of ways to choose 3 distinct primes:
inom{15}{3} = rac{15 \cdot 14 \cdot 13}{6} = 455
Number of favorable outcomes (include 2 and two odd primes):
Choose 2 from the 14 odd primes:
inom{14}{2} = rac{14 \cdot 13}{2} = 91

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