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 and E_3` occurs satisfy the equations `(alpha-2beta), p=alphabeta and (beta-3gamma) p=2beta gamma.` All the given probabilities are assumed to lie in the interval `(0,1).` Then,`(probability of occurrence of E_1) / (probability of occurrence of E_3)` is equal to

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.`