Question

If A, B, C are mutually exclusive and exhaustive events, then \[P(A) + P(B) + P(C)\] equals to
A) \[\dfrac{1}{3}\]
B) 1
C) 0
D) Any value between 0 and 1

Answer 1

Hint: When two events cannot occur at the same time, then we say that the events are mutually exclusive else the events are said to be not mutually exclusive.Two or more events are said to be exhaustive if there is a certain chance of occurrence of at least one of them when they are all considered together.
Complete Step by Step Solution:
For A, B and C to be 3 events and P(A), P(B) and P(C) be the three probabilities for the events to take place
Then the points to be noted are
I) The maximum possible probability is 1, which is the probability of a sure event.
II) Since P(A), P(B) and P(C) are exhaustive they are the only three events.
III) \[P(A \cup B \cup C) = 1;\] where \[A \cup B \cup C\] is a sure event as one of the two events are sure to occur for the experiment.
Therefore we can write that \[P(A) + P(B) + P(C) = 1\] which makes option B to be the correct option here.

Note: When two events are only mutually exclusive then only their probabilities are normally added then it can take any value from 0 to 1.