Let there be three independent events E1,E2 and E3. The probability that only E1 occurs  is α, only E2 occurs is β and only E3 occurs is γ . Let 'p' denote the probability of none  of events occurs that satisfies the equations (α-2β)p=αβ and (β-3γ)p=2βγ . All  the given probabilities are assumed to lie in the interval (0,1) .  Then,  Probabilityof occurrence of E1 Probability of occurrence of E3 is equal to

Suppose that the probabilities of happening the events E1,E2,E3  are x,y,z respectivelyHence, PE1=x,PE2=y,PE3=zSince α  is probability that the event  only E1 occursit means α=x1−y1−z,  it gives PE1⋅PE2'⋅PE3'similarly β=1−xy1−z,γ=1−x1−yzp=PE1'⋅PE2'⋅PE3'=1−x1−y1−zSubstitute the above expressions in the given relationsα−2βp=αβx1−y1−z−21−xy1−z1−x1−y1−z=xy1−x1−y1−z2Cancel out the common termsx1−y−21−xy=xyx−xy−2y+2xy=xyx−2y=0                        ......1Use the other relation β−3γp=βγwe get y=3z                .......2From the above equations, x=6zHence, PE1PE3=xz=6zz=6