Let `E` be an event which is neither a certainty nor an impossibility. If probability is such that `P(E)=1+lambda+lambda^2` and `P(E^(prime))=(1+lambda)^2` in terms of an unknown `lambdadot` Then `P(E)` is equal to `1` b. `3//4` c. `1//4` d. none of these

Options

A. 1
B. `(3)/(4)`
C. `(1)/(4)`
D. none of these

Answer

B

Explanation

P(E) + P(E') = 1
`therefore 1 + lambda + lambda^(2) + (1 + lambda)^(2) = 1`
`implies 2lambda^(2) + 3lambda + 1 = 0`
`implies (2lamda + 1)(lambda + 1) = 0`
`implies lambda = - 1, - (1)/(2)`
For `lambda = -1, P(E) = 1 + (-1) + (-1)^(2) = 1` (not possible)
For `lambda = - (1)/(2)`, P(E) = 1 + `(-(1)/(2)) + (-(1)/(2))^(2) = (1)/(2) + (1)/(4) = (3)/(4)`