Suppose you have a coin with 4 fair sides, flip it 5 times, and want to know the probability of 5 heads. This is(adsbygoogle = window.adsbygoogle || []).push({});

K(10,5) * (0.25)^{5}* (1-0.25)^{5}= K(10,5)*0.25^{5}*0.75^{5}

Or more generally for any binomially distributed outcome:

1) p(x=r) = p^{r}*(1-p)^{n-r}*K(n,r)

But also we must have that:

2) p(x=r) = K(n,r)/total combinations = K(n,r)/4^{n}

How do you show that 1) and 2) are equivalent?

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Binomial distribution

Loading...

Similar Threads - Binomial distribution | Date |
---|---|

Radioactive decay, relation between binomial to expon. dist | Jul 9, 2016 |

Binomial Distribution for successive events | Nov 26, 2015 |

Comparing probabilities of Binomially-distributed RVs | Sep 21, 2015 |

Chernoff Bound for Binomial Distribution | Jul 31, 2015 |

**Physics Forums - The Fusion of Science and Community**