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What is the probability of getting a second head on the 5th trial of 5 independent trials of tossing a fair coin? Hint use negative binomial

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In order for this to happen, you will have to throw heads once during the first four trials.

The probability of this happening is easy to calculate: there are 2^5 possible sequences (two options for every trial, and five trials), And there are four with just one heads in the first four (You can throw a head the first, second third or fourth trial).

No matter when you threw the one heads in the first four trial, there is only one possible way of getting heads in the fifth. So for every way you get heads in the first four trials, there is one way of getting heads in the fifth. This is 4*1 = 4

These are all the good outcomes. The good outcomes divided by the total outcomes is the probability. So 4/2^5 = 4/32 = 1/8 ----> 12.5% or 0.125

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Hint: You'd need exactly one head in the first four tosses, and then a head on the fifth toss.

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