Question

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

Answer 1

PLAN
For the events to be independent,
`P(E_(1)nnE_(2)nnE_(3))=P(E_(1)).P(E_(2)).P(E_(3))`
`P(E_(1)nnE_(2)nnE_(3))=P("only "E_(1)" occurs")`
`=P(E_(1)).(1-P(E_(2)))(1-P(E_(3)))`
Let x,y and z be probabilites of `E_(1),E_(2)" and "E_(3)`,respectivey.
`:." "alpha=x(1-y)(1-z)" "...(i)`
`" "beta=(1-x).y(1-z)" "...(ii)`
`" "gamma=(1-x)(1-y)z" "...(iii)`
`implies" "p=(1-x)(1-y)(1-z)" "...(iv)`
Given, `(alpha-2beta)p=alphabeta" and "(beta-3gamma)p=2betagamma" "...(v)`
From above equations, `x=2y" and "x=3z`
`:." "x=6z`
`implies" "(x)/(z)=6`