Conditional Binomial Distribution

Thread starter shespuzzling
Start date Oct 27, 2010
Tags

Binomial Binomial distribution Conditional Distribution

AI Thread Summary

To find a conditional binomial distribution, the formula F(k=7|k >= 4) = P(k=7, k>=4)/P(k>=4) is used. This requires calculating the joint probability P(k=7, k>=4) and the marginal probability P(k>=4). The discussion clarifies that P(k=7, k>=4) is equal to P(k=7) since k=7 inherently satisfies k>=4. Thus, the conditional probability simplifies to F(k=7|k >= 4) = P(k=7)/P(k>=4). Understanding these probabilities is essential for correctly applying the conditional binomial distribution in experiments.

Oct 27, 2010

shespuzzling

How do I find a conditional bionomial distribution? For example, if I want the probability that k=7 (for instance, 7 could be any number depending on the experiment), given that k is greater/equal to 4. I know what the equation would look like

i.e.: F(k=7|k >= 4)= P(k=7, k>=4)/P(k>=4). Then, would this be equal to P(k=7)/P(k>=4)? Thanks in advance for your help.

Physics news on Phys.org

Oct 27, 2010

statdad

Homework Helper

Yup.

Oct 27, 2010

shespuzzling

thanks!

Thread 'Value of intuitionistic logic'

I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.

View full post »

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

View full post »