muIf A, B and C are three mutually exclusive events, the probability of their union is equal to _____.

P(A) P(B) P(C)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P(A) + P(B) + P(C) - P(ABC)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P(A) + P(B) + P(C)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P(A) P(B) + P(C) P(A) + P(B) P(C)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C P(A) + P(B) + P(C)
If A,B and C are three mutually exclusive events then there will be no common points between these events, so there will be no intersection of these events. So P(AUBUC) = P(A)+P(B)+P(C).

Suggest Corrections

Similar questions

P (A) = \frac{1}{4}

P (B) = \frac{1}{6}

P (C) = \frac{2}{3}

P (A \cup B \cup C) = P (A) + P (B) + P (C)

A, B

C

P (A) + P (B) + P (C)

P (A) = \frac{2}{3}, P (B) = \frac{1}{4}, P (C) = \frac{1}{6} .

A, B, C

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (b \cap C) - P (C \cap A) + P (A \cap B \cap C)

A, B, C

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)

Join BYJU'S Learning Program