# Solve the following: For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = 23, P(B ∪ C) = 34, P(A ∪ B ∪ C) = 1112. F - Mathematics and Statistics

Sum

Solve the following:

For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = 2/3, P(B ∪ C) = 3/4, P(A ∪ B ∪ C) = 11/12. Find P(A), P(B) and P(C)

#### Solution

It is given that

P(A ∪ C) = 2/3, P(B ∪ C) = 3/4, P(A ∪ B ∪ C) = 11/12.

P(A ∪ C) = 2/3 gives,

P(A) + P(C) – P(A ∩ C) = 2/3  ...(1)

P(B ∪ C) = 3/4 gives,

P(B) + P(C) – P(B ∩ C) = 3/4  ...(2)

P(A ∪ B ∪ C) = 11/12 gives,

P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C) = 11/12

∴ P(A) + P(B) + P(C) – P(A n C) – P(B n C) = 11/12  ...[(because "A""," "B"  "are disjoint"),(therefore "A" ∩  "B" = "A" ∩  "B" ∩  "C" = phi)]

∴ "P"("A") + "P"("B") + "P"("C") – ["P"("A") + "P"("C") - 2/3] - ["P"("B") + "P"("C") - 3/4] = 11/12  ...[By (1) and (2)]

∴ −P(C) = 11/12 - 2/3 - 3/4 = -1/2

∴  P(C) = 1/2

From (1),

P(A) + P(C) – P(A)·P(C) = 2/3  ...[∵ A, C are independent]

∴ "P"("A") + 1/2 - 1/2"P"("A") = 2/3

∴ 1/2"P"("A") = 1/6

∴ P(A) = 1/3

From (2),

P(B) + P(C) – P(B) P(C) = 3/4  ...[∵ B, C are independent]

∴ "P"("B") + 1/2 - 1/2 "P"("B") = 3/4

∴ 1/2"P"("B") = 1/4

∴ P(B) = 1/2

∴ P(A) = 1/3, P(B) = P(C) = 1/2

Concept: Independent Events
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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 9 Probability
Miscellaneous Exercise 9 | Q II. (16) | Page 214