Re: probability of coin toss
- From: "Luis A. Afonso" <licas_@xxxxxxxxxxx>
- Date: Fri, 28 Dec 2007 12:17:34 EST
Nobody wrote
*** Date: Dec 27, 2007 5:51 PM
Author: nobody
Subject: probability of coin toss
I am having a difficult time trying to figure what is the best method to calculate the probability of a number of unbiased coin tosses.
Example: What is the probability of tossing 90 heads out of 100 coin tosses?
Any insight would be greatly appreciated!
Nick ***
My response
1110, 1101, 1011, 0111
are the four ways to have 3 heads (1) in four coin tosses. If p is the probability to head up then q=1-p is the tails (0) probability.
SINCE THAT THEY ARE INDEPENDENT TRIALS then the probability to occur
____1110_____is___ p*p*p*q
____1101_________ p*p*q*p
____1011_________ p*q*p*p
____0111_________ q*p*p*p
EACH ONE is (p^3)*q which is (p^x)*[q^(n-x)] with x=3 and n=4.
How many groups can we make with n objects, each group containing x of the class 1 without repetition (each object chosen ONCE)? .
Answer: n Choose x:
__________ nCx = n! /[x! * (n-x)!].
Luis Amaral Afonso
.
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