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Your answer is right. The solution can be derived using Bayes' Theorem : $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ You want to know the probability of $P...

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In general, P(biased coin|k heads) = (2^k)/[(2^k) + 99] Where k is no of consective heads so if the trick coin was tossed 3 times (2^3)/[(2^3) + 99...

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How do you find the probability of a coin being biased?

To find the probability of anything other than all heads, simply note that 'all heads + anything else = 1'. For the fair coin then, there's a 1 - (0.5^n) chance of getting anything other than all heads and that chance is 1 - (0.75^n) for the biased coin.

What is the probability of getting all heads on a coin?

A fair coin has a 50% chance of landing on heads or tails. A fair coin flipped n times has a (0.5^n) chance of getting all heads. Likewise, our biased coin flipped n times has a (0.75^n) chance of getting all heads. To find the probability of anything other than all heads, simply note that 'all heads + anything else = 1'.

Are most coins fair or biased?

Most coins are close to being fair. That is, they have the same probability of landing heads and tails (0.5). You can talk about the bias in terms of only one of those probabilities. There are only two possible outcomes and the other probability has to be 1 minus the first.

What is the probability of all three tosses of a coin?

The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. The probability of three heads given the biased coin is trivial: P ( three heads | biased coin) = 1. P ( biased coin | three heads) = 1 × 1 / 100 1 + 99 × 1 8 100 = 1 1 + 99 × 1 8 = 8 107 ≈ 0.07476636

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