Question

Let $A, B, C$ be all the three possible mutually exclusive events of a random experiment. Which one of the following is not permissible in terms of their probabilities?

• A. $P(A)=\frac{7}{19}, \; P(B)=\frac{4}{19}, \; P(C)=\frac{8}{19}$
• B. $P(A)=\frac{18}{95}, \; P(B)=\frac{29}{95}, \; P(C)=\frac{48}{95}$
• C. $P(A)=\frac{81}{190}, \; P(B)=\frac{41}{190}, \; P(C)=\frac{68}{190}$
• D. $P(A)=\frac{21}{95}, \; P(B)=\frac{42}{95}, \; P(C)=\frac{32}{95}$
• E. $P(A)=\frac{77}{190}, \; P(B)=\frac{47}{190}, \; P(C)=\frac{67}{190}$

Hint:

Sum of probabilities of all possible outcomes must always be 1.

Answer 1

Answer: (E)

Concept:
• For mutually exclusive and exhaustive events: \[ P(A)+P(B)+P(C) = 1 \]

Step 1: Check Option (E)
\[ \frac{77}{190} + \frac{47}{190} + \frac{67}{190} = \frac{191}{190}>1 \]

Step 2: Conclusion
Sum exceeds 1, which is not possible.

Step 3: Check others (brief)
All other options sum to exactly 1 $\Rightarrow$ valid. Final Conclusion:
Option (E) is not permissible.