Question

Let P(x) denotes the probability of the occurrence of event x.

Then all those points (x, y) = (P(1), P(2)), in a plane which

satisfy the conditions,

P (A Č B) ³$\frac { 1 } { 8 }$£ P (A Ē B) £ $\frac { 3 } { 8 }$ implies

• A. P (1) + P (2) <$\frac { 11 } { 8 }$
• B. P (1) + P (2) > $\frac { 11 } { 8 }$
• C. $\frac { 7 } { 8 }$£ P (1) + P (2) £$\frac { 11 } { 8 }$
• D. None of these

Answer 1

Answer: (C)

$\frac { 7 } { 8 }$£ P (1) + P (2) £$\frac { 11 } { 8 }$

Answer 2

P (A Č B) ³ $\frac { 3 } { 8 }$.

Ž P (1) + P (2) – P (A Ē B) ³ $\frac { 3 } { 4 }$

Ž P (1) + P (2) ³ $\frac { 3 } { 4 }$ + P (A Ē B) = $\frac { 3 } { 4 }$ + $\frac { 1 } { 8 }$ = $\frac { 7 } { 8 }$

We know that P (A Č B) £ 1

P (1) + P (2) £ 1 + P (A Ē B) £ 1 + $\frac { 3 } { 8 }$ = Ž x + y £ $\frac { 11 } { 8 }$

Ž $\frac { 7 } { 8 }$ £ x + y £ $\frac { 11 } { 8 }$

Also 0 £ P (1) £ 1, 0 £ P (2) £ 1 .

Hence (3) is the correct answer