- π(π=π₯; π,π) = π{ the π₯π‘β Bernoulli trial results in the ππ‘βΒ success }
- π(π=π₯; π,π) = π{ (π-1) successes in the first (π₯-1) Bernoulli trials AND the last Bernoulli trial is success }
- π(π=π₯; π,π) =Β π{ (π-1) successes in the first (π₯-1) trials }Β π{ the last trial is success }
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π(π=π₯; π,π) = π(π=π-1; π,π=π₯-1) π(π=π₯) # where:
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- <font style="color: rgb(0,128,0);"><font style="color: rgb(255,102,0);">π(π=π-1; π,π=π₯-1)<font style="color: rgb(122,134,154);">Β is a Binomial Distribution</font></font></font>
- <font style="color: rgb(0,128,0);">π(π=π₯)<font style="color: rgb(122,134,154);">Β is a Bernoulli Distribution</font></font>
- π(π=π₯; π,π) =Β [(π₯-1) choose (π-1)]Β (1-π)π₯-πππ-1π
- π(π=π₯; π,π) =Β [(π₯-1)Β chooseΒ (π-1)]Β (1-π)π₯-πππ
- π(π=π₯; π,π) =Β [(π₯-1)!/[(π-1)!(π₯-π)!]]Β (1-π)π₯-πππ
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- π(π=π₯; π,π) =Β π{ exactly π₯ success in π Bernoulli trialsΒ }
- π(π=π₯; π,π) = [πΒ chooseΒ π₯] * prob-of-π₯-success * prob-of-(π-π₯)-failures
- π(π=π₯; π,π) = [π!/(π₯!(π-π₯)!] (1-π)π-π₯ππ₯
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Let π=π
- π(π=π; π,π) = [(π-1)!/[(π-1)!(π-π)!]] (1-π)π-πππ
- π(π=π; π,π) = (π/π) π(π=π; π,π) # substitute Binomial Distribution
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Let π=π
- π(π=π; π,π) = [π!/(π!(π-π)!]Β (1-π)π-πππ
- π(π=π; π,π) = (π/π)Β π(π=π; π,π) # substitute Negative Binomial Distribution
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Example Bernoulli trials for NBD π₯=10 and π=4:
- π² β¬οΈ π² β¬οΈ π² β¬οΈ π² π² π² β¬οΈ
- π² π² β¬οΈ β¬οΈ π² π² β¬οΈ π² π² β¬οΈ
- β¬οΈ π² π² β¬οΈ β¬οΈ π² π² π² π² β¬οΈ
- π² π² β¬οΈ β¬οΈ β¬οΈ π² π² π² π² β¬οΈ
NOTE: that every tenth box is black (e.g. success)
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Example Bernoulli trials for BD π=10 and π₯=4:
- π² β¬οΈ π² β¬οΈ π² β¬οΈ π² π² β¬οΈ π²
- π² π² β¬οΈ β¬οΈ π² π² β¬οΈ π² π² β¬οΈ
- β¬οΈ π² π² β¬οΈ β¬οΈ π² π² π² β¬οΈ π²
- π² π² β¬οΈ β¬οΈ β¬οΈ β¬οΈ π² π² π² π²
NOTE: that the tenth box doesnβt need to be black (e.g. success)
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