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01-22 Probability Review Pt 3 Solution

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    We can compute this using the formula for conditional probability.
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    In this case what A is is the probability the coin toss is valid.
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    A is valid, and we know that the probability of A is equal to 0.99998.
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    What B is is the probability that it's heads.
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    So the probability of B is the probability of H, which is 0.49999.
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    Now we just have to plug these into the formula.
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    But what we need to use the formula is the probability of A intersect B.
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    And what A is the probability of A intersect B. What A is is valid.
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    Instead of heads and tails, what B is is heads.
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    If we intersect heads and tails with heads, we get tails.
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    We know the probability of tails is 0.49999.
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    That means using the formula we have the probability of tails, which is 0.49999,
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    which is the probability of A intersect B.
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    We're dividing that by the probability of A,
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    which is a valid event which is 0.99998.
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    We get 0.5.
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    I should note that this is not the case for real coin tosses.
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    There is no physical coin ever manufactured that has exactly
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    chances of landing on both sides.
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    In fact, with real coin tosses, at least with American currency,
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    there is a much higher percentage--much higher meaning close to 51%
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    rather than 50%--that the coin lands on the same side that it started on.
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    When we talk about mathematical coin tosses,
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    we're going to assume that there is no edge case
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    and that it's equally likely
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    that we have a uniform distribution and there are only 2 outcomes.
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    When we talk about mathematical coin tosses,
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    we're going to assume that we have a uniform distribution,
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    and there are only two outcomes.
Title:
01-22 Probability Review Pt 3 Solution
Description:

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Team:
Udacity
Project:
CS387 - Applied Cryptography
Duration:
02:01
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