Question

If $\left( \frac { 1 + a } { 3 } \right)$ and $\left( \frac { 1 - \mathrm { a } } { 4 } \right)$ are probability of two mutually exclusive events, then set of all values of a is:

• A. -1 ≤ a ≤ 1
• B.
  • 7 ≤ a ≤ 5
• C.
  • 1 ≤ a ≤ 2
• D.
  • 4 ≤ a ≤ 1

Answer 1

Answer: (A)

-1 ≤ a ≤ 1

Answer 2

Since, 0 ≤ ⇒ -1 ≤ a ≤ 2

and 0 ≤ $\frac { 1 - \mathrm { a } } { 4 }$ ≤ 1 ⇒ -3 ≤ a ≤ 1

Also, as $\frac { 1 + \mathrm { a } } { 3 }$ and $\frac { 1 - \mathrm { a } } { 4 }$ are the probabilities of two

mutually exclusive events.

0 ≤ $\frac { 1 + \mathrm { a } } { 3 } + \frac { 1 - \mathrm { a } } { 4 }$ ≤ 1 ⇒ -7 ≤ a ≤ 5.

From equation (i), (ii) and (iii), we get - 1 ≤ a ≤ 1.