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You are given a bag of coins with varying biases.the probability of heads is a random variable sample from uniform distribution u[0,1].you draw a coin from these bag and toss it 100 times what is the probability of getting 100 heads stackover flow

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Final answer:

The probability of getting 100 heads in 100 tosses of a biased coin drawn from a bag with biases uniformly distributed between 0 and 1 requires integrating over all possible biases. Results from fair coin tosses hint at the improbability of this outcome according to the law of large numbers. Exact probabilities from Table 15.4 confirm the rarity of such an event.

Step-by-step explanation:

The student is asking about the probability of getting 100 heads when a randomly selected biased coin is tossed 100 times. The bias of the coin is described by a random variable from a uniform distribution between 0 and 1. To answer this question, we must consider the variable probability of the coin landing on heads. Each biased coin could have a different probability of landing heads, but without knowing the exact bias of the drawn coin, we calculate an overall probability of getting 100 heads in 100 tosses by integrating over all possible biases.

However, the provided information from other contexts gives us insights into probability and expected outcomes but does not directly answer the question about the biased coin. We do know that in large numbers of coin tosses, a fair coin will tend towards a 50/50 distribution of heads and tails due to the law of large numbers.

From Table 15.4, we can infer that getting 100 heads in 100 tosses is very unlikely, as the number of microstates for such an event is extremely low compared to other more balanced outcomes.

User Janetsmith
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Answer:


Step-by-step explanation:




Suppose I have a bag of n biased coins, and I sample without replacement m<n of them, and measure each coin i for their parameter pi∈[0,1], that is each coin is Bernoulli(pi). Now I would like to ask what is the most likely pm+1 on the next coin I pick. I am not sure how to answer this question aside from taking the mean of all m coins' parameter so far. That is: p^m+1=p1+…+pmm.

User Karthik H
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