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Product of Covariance Matrices
Concise Answer: No, generally AB is not a covariance matrix; however, if AB=BA, then AB could be a covariance matrix.Detailed Explanation: The product of two covariance matrices does not necessarily yield another covariance matrix because it may not satisfy properties like symmetry or non-negative dRead more
Concise Answer: No, generally AB is not a covariance matrix; however, if AB=BA, then AB could be a covariance matrix.
See lessDetailed Explanation: The product of two covariance matrices does not necessarily yield another covariance matrix because it may not satisfy properties like symmetry or non-negative definiteness. If AB=BA, it suggests commutativity, which is a property of covariance matrices for joint Gaussian distributions, hence under specific conditions, AB could be a covariance matrix.
Data Sort
Concise Answer: Implement an external merge sort.Detailed Explanation: Divide the data into 1 GB chunks and sort each chunk in memory. Then perform a k-way merge on the sorted chunks, reading and merging chunks into larger sorted segments until the full dataset is sorted.
Concise Answer: Implement an external merge sort.
See lessDetailed Explanation: Divide the data into 1 GB chunks and sort each chunk in memory. Then perform a k-way merge on the sorted chunks, reading and merging chunks into larger sorted segments until the full dataset is sorted.
Red Marble Optimal Draw
Concise Answer: Place one red marble in one drawer and all other marbles in the second drawer.Detailed Explanation: To maximize the chance of drawing a red marble, you can exploit the fact that selecting the drawer and the marble are sequential events. By placing one red marble alone in one drawer,Read more
Concise Answer: Place one red marble in one drawer and all other marbles in the second drawer.
See lessDetailed Explanation: To maximize the chance of drawing a red marble, you can exploit the fact that selecting the drawer and the marble are sequential events. By placing one red marble alone in one drawer, the probability of drawing it (if that drawer is chosen) is 100%. Since drawers are selected randomly, the overall probability of getting a red marble is maximized.
Biased Coin
Concise Answer (continued): Yes, use a sequence of flips to simulate a fair outcome.Detailed Explanation (continued): For an unfair coin with any bias, you can flip it multiple times to create pairs of outcomes such as HT and TH. If the coin lands HT, you decide "heads," and if it lands TH, you deciRead more
Concise Answer (continued): Yes, use a sequence of flips to simulate a fair outcome.
Detailed Explanation (continued): For an unfair coin with any bias, you can flip it multiple times to create pairs of outcomes such as HT and TH. If the coin lands HT, you decide “heads,” and if it lands TH, you decide “tails.” If it lands HH or TT, you disregard the outcome and flip again. This method effectively ‘cancels out’ the bias, as HT and TH have the same probability, and you keep flipping until you get one of these combinations. This process ensures a 50% chance for each outcome and is known as the von Neumann extractor for fairness.
Now, let’s add more detail for parts b), c), and d) since these require a deeper explanation:
b) Expected Number of Flips (Detailed): The expected number of flips until the realization of an event is the inverse of the event’s probability. If the event is “getting heads” with P(H) = 1/3, the expected number of flips is 1/(1/3), which equals 3 flips. For two consecutive heads, we need to consider the geometric series formed by the probabilities of getting two heads in a row for each series of independent flips. The probability of getting two heads in a row is (1/3)^2 = 1/9, so the expected number of flips would be 1/(1/9) = 9 flips. However, because any flip sequence ending in “TH” resets the count for two consecutive heads, the calculation is a bit more complex, resulting in an expectation of 12 flips.
c) Reducing the Number of Flips (Concise): No strategy can change the expected number of flips because the coin’s bias is constant, and each flip is independent.
c) Reducing the Number of Flips (Detailed): Since the outcomes are determined by the biased probability of the coin, the expected number of flips for a given event cannot be changed without changing the underlying probability. However, what can be done is optimizing the process by avoiding unnecessary flips once the desired event has occurred or by using outcomes in combination to decide an event (as in the HT/TH method).
d) Strategy for Any Unfair Coin (Concise): The HT/TH method can be adapted for any bias.
d) Strategy for Any Unfair Coin (Detailed): The general approach to create a fair outcome from a biased coin is to flip the coin twice, creating four outcomes: HH, HT, TH, and TT. For any bias, the outcomes HT and TH will have the same probability, which allows us to use them to simulate a fair coin. If neither of these outcomes occurs, the flips are repeated. This strategy ensures a fair event from a biased coin with any probability distribution, as long as the coin is not deterministic (i.e., P(H) != 0 or 1).
See lessGirlfriend Dinning
Concise Answer: The train to your girlfriend's place in uptown must be arriving more frequently.Detailed Explanation: If both trains run at the same intervals and you're taking the first one that arrives without bias, theoretically, you should end up dining with both equally. If you're ending up witRead more
Concise Answer: The train to your girlfriend’s place in uptown must be arriving more frequently.
See lessDetailed Explanation: If both trains run at the same intervals and you’re taking the first one that arrives without bias, theoretically, you should end up dining with both equally. If you’re ending up with your girlfriend 90% of the time, it suggests that the train to uptown arrives three times as often as the one to downtown during the times you go to the station.
Coin Flip
Concise Answer: The probability is 1/3.Detailed Explanation: There are three equally likely scenarios where you get at least one head in three flips: HHT, HTH, and THH. Since one head is certain, only these three outcomes are possible, making the probability of two heads and one tail 1/3.
Concise Answer: The probability is 1/3.
See lessDetailed Explanation: There are three equally likely scenarios where you get at least one head in three flips: HHT, HTH, and THH. Since one head is certain, only these three outcomes are possible, making the probability of two heads and one tail 1/3.
Concise Answer: The average speed is 133.33 mph.Detailed Explanation: The average speed is not simply (100 + 200) / 2 = 150 mph because average speed is total distance divided by total time. The first round at 100 mph and the second at 200 mph each cover the same distance, so if the distance is D, tRead more
Concise Answer: The average speed is 133.33 mph.
See lessDetailed Explanation: The average speed is not simply (100 + 200) / 2 = 150 mph because average speed is total distance divided by total time. The first round at 100 mph and the second at 200 mph each cover the same distance, so if the distance is D, the total distance is 2D, and the time taken is D/100 + D/200. The average speed is 2D / (D/100 + D/200) = 133.33 mph.
12 Ball Weighting
Concise Answer: Divide and balance.Detailed Explanation: Divide the balls into three groups of four. Use the scale to compare two groups. If they balance, the different ball is in the third group. If not, it's in the heavier group. Then, from the identified group, place two balls on each side of theRead more
Concise Answer: Divide and balance.
See lessDetailed Explanation: Divide the balls into three groups of four. Use the scale to compare two groups. If they balance, the different ball is in the third group. If not, it’s in the heavier group. Then, from the identified group, place two balls on each side of the scale. If they balance, the remaining ball is the different one. If not, you’ve identified the heavier ball. You can determine which ball is different in the third weighing.