Using Chebyshev's Theorem (and another minor question)

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Chebyshev's Theorem can be applied to find an upper bound for P(X = 21) by determining a suitable value for k, with k being greater than 3 based on the variance of 9. The inequality P(|21 - 12| < 3k) ≤ 1/k² is used to establish this bound. For the second question regarding events A and B, the relationship between their probabilities suggests that the intersection must fall between 1/12 and 1/3 due to the total probability exceeding 1. The calculations involving P(A U B) indicate that the overlap between A and B must be considered to understand the bounds of their intersection. Overall, the discussion emphasizes the application of probability theory to establish bounds and relationships between events.
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Homework Statement



If X is a discrete random variable with mean u = 12 and variance = 9, use Chebyshev's Theorem to find an upper bound for P(X = 21).

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The Attempt at a Solution



Now, I'm not sure about this since there are different upper bounds, right?

P(|21 - 3k < 21 < 21 + 3k|) ≤ 1 / k^2. We solve the inner inequality and we get -k < 3 < k. To find an upper bound, do we simply take a value k > 3?

And an unrelated question (just not to make another thread):

Let A and B be two events with P(A) = 3/4 and P(B) = 1/3. Explain why 1/12 ≤ P(A intersection B) ≤ 1/3. How do we approach this exactly?
 
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regarding your second question, just something to think about: what is P(A Union B)? what does this tell you about P(A intersect B)?
 
P(A U B) = P(A) + P(B) - P(A intersection B) = 3/4 + 1/3 - P(A intersection B) = 13/12 - P(A intersection B). So, when P(A intersection B) = 1/12, P(A U B) = 1 and when it's 1/3, P(A U B) = 3/4. Does it have to do something with that? It looks too obvious but I'm not figuring it out!
 
the fact that P(A) + P(B) > 1 tells you there has to be some overlap, right? So you're looking at kind of a best case/worst case scenario. How small could the overlap be? How large could the overlap be?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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