# If ( 1 − 3 P ) 2 , ( 1 + 4 P ) 3 , ( 1 + P ) 6 Are the Probabilities of Three Mutually Exclusive and Exhaustive Events, Then the Set of All Values of P is - Mathematics

MCQ

If $\frac{(1 - 3p)}{2}, \frac{(1 + 4p)}{3}, \frac{(1 + p)}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is

• (0, 1)

•  (−1/4, 1/3)

• (0, 1/3)

• (0, ∞)

#### Solution

(−1/4, 1/3)

P(A) = (1 − 3p)/2
P(B) = (1 + 4p)/3
P(C) = (1 + p)/6

The events are mutually exclusive and exhaustive.
∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1

⇒$0 \leq P\left( A \right) \leq 1$

$0 \leq P\left( B \right) \leq 1$
$0 \leq P\left ( C \right) \leq 1$
⇒ $0 \leq \frac{1 - 3p}{2} \leq 1$

$0 \leq \frac{1 + 4p}{3} \leq 1$

$0 \leq \frac{1 + p}{6} \leq 1$

$\Rightarrow - 1/3 \leq p \leq \frac{1}{3}$ ,      ...(i)

$\frac{- 1}{4} \leq p \leq \frac{1}{2}$     ...(ii)

and $- 1 \leq p \leq 5$       ...(iii)
The common solution of (i), (ii) and (iii) is$- 1/4 \leq p \leq 1/3$
∴ The set values of p are (- 1/4 , 1/3)

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 33 Probability
Q 14 | Page 72

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