Question

Assume that a procedure yields a binomial distribution with a
trial repeated n=5 times. Find the probability distribution given
the probability p=0.533 of success on a single trial.

*(Report answers accurate to 4 decimal places.)*

k |
P(X = k) |
---|---|

0 | |

1 | |

2 | |

3 | |

4 | |

5 |

Answer #1

Here, n = 5, p = 0.533, (1 - p) = 0.467 and x = 0

As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)

We need to calculate P(X = 0)

P(X = 0) = 5C0 * 0.533^0 * 0.467^5

P(X = 0) = 0.0222

P(X = 1) = 5C1 * 0.533^1 * 0.467^4

P(X = 1) = 0.1268

P(X = 2) = 5C2 * 0.533^2 * 0.467^3

P(X = 2) = 0.2893

P(X = 3) = 5C3 * 0.533^3 * 0.467^2

P(X = 3) = 0.3302

P(X = 4) = 5C4 * 0.533^4 * 0.467^1

P(X = 4) = 0.1884

P(X = 5) = 5C5 * 0.533^5 * 0.467^0

P(X = 5) = 0.0430

k | P(X = k) |

0 | 0.0222 |

1 | 0.1268 |

2 | 0.2893 |

3 | 0.3302 |

4 | 0.1884 |

5 | 0.0430 |

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