Recent content by rangatudugala

Big Thanks Sir... Now I got the Point we don't know anything about event a & b so we can do both prove & disprove using "if statements. Thank you sir again. Big sorry if I bothered to you so much.. :)

1/ let a, b are independent event 0<q<1 then p(a∩b) = p(a) p(b) = q*q = q^2 2/ a, b not independent event 0<q<1 then p(a∩a) = q >= q^2but why we going to calculate p(a∩a)?? question was p(a∩b) =< q^2 know i confused again.. Am I missed step ?

p(a)=p(a) =q yes !

its greater than q^2

isnt it?

p(a ∩ a) = 0.5 :?

p(a∩b) = p(a) + p(b) - p(a∪b)Then how to get p(a∪b)??

is there any other answer ??

Okay.. then p(a∩b) = p(a) p(b) =0.5*0.5 = 0.25

I tried but no idea feel something missing let a, b are independent event 0<q<1 then p(a∩b) = p(a) p(b) = q*q = q^2

previous thread I used confusing notation (It was my 1st attempt )

Prove or disprove the following statement: If p(a)=p(b)=q then p(a∩b)≤q2 We know nothing know about event a , b. The Attempt at a Solution I tried this but don't know correct or not Can some one help me let a, b are independent event 0<q<1 then p(a∩b) = p(a) p(b) = q*q = q^2 [/B]

I hv already answered. probably you didn't see here rewritten the question prove or disprove the following p(a) = P(b) = p (let say some valve, you can use what ever symbol ) then p(a ∩ b) ≤ p^2

P(a/b) means probability of a given b, right? yes its true how come this possible ?P(a/b) = P(a ∩ b) / P(b)