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=betagamma.` 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))=______.`

Options

A. 3
B. 2
C. 6
D. 4

Answer

C

Explanation

Let `P(E_(1))=x, P(E_(2))=y " and "P(E_(3))=z`
It is given that : P(Only `E_(1)` occurs)`=alpha`
`implies P(E_(1) cap overline(E_(2))capoverline(E_(3)))=alpha`
`implies P(E_(1)) P(overline(E_(2)))P(E_(3))=alpha`
`implies x(1-y)(1-z)=alpha`
P(Only `E_(2)` occurs)`=beta`
`implies P(overline(E_(1)) cap E_(2) cap overline(E_(3)))=beta`
`implies P(overline(E_(1)))P(E_(2))P(overline(E_(3)))=beta`
`implies (1-x)y(1-z)=beta`
P(only `E_(3)` occurs)` = gamma`
`implies P(overline(E_(1)) cap overline(E_(2)) cap E_(3))=gamma`
`implies P(overline(E_(1))) P(overline(E_(2)))P(E_(3))=gamma`
`implies P(overline(E_(1)))P(overline(E_(2)))P(E_(3))=gamma`
`implies (1-x) (1-y) z=gamma`
and, P(None of the events `E_(1),E_(2) " and " E_(3)` occurs)=p
`implies P(overline(E_(1)) cap overline(E_(2)) cap overline(E_(3)))=p`
`implies P(overline(E_(1)))P(overline(E_(2)))P(overline(E_(3)))=p`
`implies (1-x)(1-y)(1-z)=p`
Substituting these values in `(alpha-2beta)p=alpha beta " and " (beta-3gamma)p=2beta gamma`, we obtain x=2y and y=3z. These two equations give x=6z.
Hence, `(P(E_(1)))/(P(E_(3)))=(x)/(z)=6`