If P(1) = (\frac { 1 } { 2 }), P(2) = - (\frac { 4 } { 7 }).... | MATH - USE OF PERMUTATIONS AND COMBINATIONS IN PROBABILITY | SolveeIt

If P(1) = $\frac { 1 } { 2 }$ , P(2) = - $\frac { 4 } { 7 }$ . Then the interval of P(AĒB)

[1/14, 1/3]

[1/14, ½]

[1/3, ½]

[0, ½]

Answer

[1/14, ½]

Explanation

$\frac { 4 } { 7 }$ £ P(AČB) £ 1

$\frac { 4 } { 7 }$ £ $\frac { 4 } { 7 } + \frac { 1 } { 2 } -$ P(AĒB) £ 1

0 £ $\frac { 1 } { 2 }$ - P(AĒB) £ $\frac { 3 } { 7 }$

$\frac { - 1 } { 2 }$ £ - P(AĒB) £ $\frac { - 1 } { 14 }$

$\frac { 1 } { 14 }$ £ P(AĒB) £ $\frac { - 1 } { 2 }$

\ P(AĒB) Ī $\left[ \frac { 1 } { 14 } , \frac { 1 } { 2 } \right]$ .

Question: If P(1) = \(\frac { 1 } { 2 }\), P(2) = - \(\frac { 4 } { 7 }\). Then the interval of P(AĒB)...