Question

If (1-3p)2,(1+4p)3,(1+p)6 are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is
(a) (0, 1)
(b) (−1/4, 1/3)
(c) (0, 1/3)
(d) (0, ∞)

Answer 1

Answer:

(b) (−1/4, 1/3)

Step-by-step explanation:

we have the given probabilities are greater than or equal to 0.

Since the probabilities of three mutually exclusive and exhaustive events.

Since (1- 3p)/2 > 0

        ⇒ 1- 3p > 0

         ⇒ -3p > -1

         ⇒ p < 1/ 3  --------(1)

Since (1 + 4p)/3 > 0

         ⇒ 1 + 4p > 0

         ⇒ 4p > -1  

         ⇒ p > -1 / 4  --------(2)

Since (1 + p)/6 > 0

           ⇒ 1 + p > 0

           ⇒ p > -1   --------(3)

from (1) , (2) and (3) we obtain set of values of p as (-1 /4 , 1/3).