Question

If $\frac { 1 + 3 \mathrm { p } } { 3 } , \frac { 1 - \mathrm { p } } { 4 } , \frac { 1 - 2 \mathrm { p } } { 2 }$ are the probabilities of 3 actually exclusive events, then

• A. –1/3 ≤p≤2/3
• B. -3≤p≤1
• C. 1/3≤p≤1/2
• D. 1/3≤p≤13/3

Answer 1

Answer: (C)

1/3≤p≤1/2

Answer 2

Let P(1) = $\frac { 1 + 3 p } { 3 }$ , P(2) = , P(3) = $\frac { 1 - 2 p } { 2 }$

P(AČBČC) = P(1) + P(2) + P(3)

= $\frac { 1 + 3 p } { 3 } + \frac { 1 - p } { 4 } + \frac { 1 - 2 p } { 2 } = \frac { 4 + 12 p + 3 - 3 p + 6 - 12 p } { 12 }$

= $\frac { 13 - 3 p } { 12 }$

0 £ P(1) £ 1 0 £ P(2) £ 1 0 £ P(3) £ 1 0 £

P(AČBČC) £ 1

0 £ 1+3P £ 3 0 £ 1- P £ 4 0 £ 1- 2P £ 2 0 £ 13 -3P £ 12

-1 £ 3p £ 2 -1 £ - p £ 3 -1 £ 2p £ 1 -1 £ 3p £ 13

$\frac { - 1 } { 3 } \leq \mathrm { p } \leq \frac { 2 } { 3 }$ -3 £ p £ 1 $\frac { - 1 } { 2 } \leq \mathrm { p } \leq \frac { 1 } { 2 }$ $\frac { 1 } { 3 } \leq \mathrm { p } \leq \frac { 13 } { 3 }$