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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)

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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
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