Based On Your Coin Flips, You Have Chosen Building To Rob

Based On Your Coin Flips You Have Chosen Building To Rob Although It

Based on your coin flips, you have chosen a building to rob. Although it has no electric fences and no closed-circuit cameras, the money is guarded by a gigantic hound. You possess a gem-powered gun capable of instantly putting any person or animal to sleep, but it works only half the time with a probability of 1/2. Additionally, you carry a dog whistle intended to keep any dog away, but it is claimed to work on only 96% of dogs. The probability that both the whistle and the gun will successfully work on this dog is 48%. The question is: what is the probability that you will be able to get past the dog by putting it to sleep, whistling, or both?

Paper For Above instruction

The scenario presented involves understanding combined probabilities of independent and dependent events, specifically in a context that models real-world uncertainties. The ultimate goal is to determine the likelihood that a robbin' individual can bypass a guarding dog using two different measures: a sleep-inducing gun and a dog whistle. This problem involves key probability concepts including the intersection and union of events, and how they relate to individual probabilities given in the problem.

First, let's define relevant events: let A be the event that the gun successfully puts the dog to sleep, and B be the event that the whistle successfully repels the dog. The problem states that the probability that both the gun and the whistle work on the dog simultaneously is 48%, or P(A ∩ B) = 0.48.

We know that the probability the gun works independently is P(A) = 0.5. The probability of success for the whistle is 0.96, so P(B) = 0.96. The goal is to find the probability that the dog is bypassed, which occurs if either the gun works, the whistle works, or both. Mathematically, this is represented as the probability of the union of the two events: P(A ∪ B).

From probability theory, the union of two events can be calculated using the inclusion-exclusion principle:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

Given the values: P(A) = 0.5, P(B) = 0.96, and P(A ∩ B) = 0.48, substitution yields:

P(A ∪ B) = 0.5 + 0.96 - 0.48 = 1.46 - 0.48 = 0.98.

Thus, the probability that you successfully get past the dog—by either putting it to sleep or whistling or both—is 98%. This high probability underscores the effectiveness of having two independent or partially independent means to evade guard dogs, especially when their individual success probabilities are combined optimally through the union of events.

Furthermore, the calculation assumes that the events "gun works" and "whistle works" are not mutually exclusive; they can both succeed at the same time, as indicated by the value of their intersection probability. The intersection probability (P(A ∩ B) = 0.48) suggests a certain degree of dependence or overlap, implying that when both methods are used simultaneously, their joint success is significant but less than perfect. This could be due to overlapping mechanisms or a common characteristic of the dog that affects both deterrent methods.

In the context of real-world applications, understanding how such combined probabilities work is critical in fields such as risk analysis, security planning, and strategic decision-making, where multiple measures are employed to mitigate threats. The calculation vividly illustrates that deploying multiple strategies can substantially increase the likelihood of success, as evidenced in this scenario where the probability of bypassing the dog approaches certainty at 98%.

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