If we are interested in calculating the probability that a first success occurs on the 6th trial, what do we know about the outcomes of the first 5 trials?

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If we are interested in calculating the probability that a first success occurs on the 6th trial, what do we know about the outcomes of the first 5 trials?

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Answer 1 · 2015-01-01T04:23:33

The first 5 trials must not be successes.

To calculate the probability of an event with probability $p$ $p$ first occurring on the sixth trial, one must assume that the event fails 5 times and then succeeds. Because each of the trials in a geometric distribution are independent, identically distributed binary events, we can simply multiply each of the 5 non-successes and the 1 success.

Because the probability of an event that succeeds with probability $p$ $p$ not succeeding is $(1 - p)$ $(1 - p)$ , the probability of the event first occurring on the sixth trial is equal to:

${(1 - p)}^{5} \cdot p$ $(1-p)^5*p$

In fact, the probability of the event occurring on the $n^{th}$ $n^"th"$ trial is:

${(1 - p)}^{n - 1} \cdot p$ $(1-p)^(n-1)*p$

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If we are interested in calculating the probability that a first success occurs on the 6th trial, what do we know about the outcomes of the first 5 trials?

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