Of the three independent events `E_(1),E_(2),and E_(3),` the probability that only `E_(1)` occurs is `alpha` only `E_(2)` occurs is `beta,` and only `E_(3)` occurs is `gamma.` Let the probability p that none of events `E_(1),E_(2), or E_(3)` occurs satisfy the equations `(alpha-2beta)p=alpha betaand (beta-3gamma)p=2betagamma.` All the given probabilities are assumed to lie in the interval `(0,1).` Then
`("Probability of occurrence of"E_(1))/("Probability of occurence of"E_(3))=`

Explanation

Let `P(E_(1))=x,P(E_(2))=y and P(E_(3))=z,then`
`(1-x)(1-y)(1-z)=p`
`x(1-y)(1-z)=alpha,`
`(1-x)y(1-z)=beta,`
`(1-x)(1-y)(z)=gamma`
`So, (1-x)/(x)=(p)/(alpha)=x=(alpha)/(alpha+p),`
Similarly, `z=(gamma)/(gamma+p).`
So, `(P(E_(1)))/(P(E_(3)))=((alpha)/(alpha+p))/(gamma/gamma+p)=((gamma+p)/(gamma))/((alpha+p)/(alpha))=(a+p/gamma)/(1+(p)/(alpha))`
Also given `(alphabeta)/(alpha-2beta)=p=(2betagamma)/(beta=3gamma)impliesbeta=(5alphagamma)/(alpha+4gamma)`
Substituting in given relation
`(alpha-2((5alpha gamma)/(alpha+4gamma)))p=(alphaxx5alphagamma)/(alpha+4gamma)`
`or ((p)/(gamma)+1)=6((p)/(alpha)+1)`
`or ((p)/(gamma+1))/((p)/(alpha)+1)=6.`