If A, B, C are mutually exclusive and exhaustive events associated to a random experiment, then write the value of P(A) + P(B) + P(C).
Given ![]()
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Also given ![]()
By formula we know, ![]()
Substituting the values we get
1 = P(A)+P(B)+P(C)
⇒ P(A)+P(B)+P(C) = 1
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