IMM - DTU 02405 Probability 2007-2-8 BFN/bfn Solution for exercise 1.3.12 in Pitman We first recall that a proof by mathematical induction includes two steps
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IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32.
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32.
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
P (B ∩ An+1)
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1))
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n
i=1Bi) ,
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n
i=1Bi) ,
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n
i=1Bi) ,
implying that we can use the inclusion-exclusion formula for this term too.
IMM - DTU 02405 Probability
2007-2-8BFN/bfn
Solution for exercise 1.3.12 in Pitman
We first recall that a proof by mathematical induction includes two steps
1. Prove that the stated formula is true for some n0
2. Prove that if the formula is true for some n then this implies, that theformula is also true for n + 1.
We know from exercise 1.3.11 that the formula is valid for n = 3 and consider
P(
∪n+1
i=1 Ai
)
= P ((∪ni=1Ai) ∪ An+1) . (1)
Using exclusion-inclusion for two events we get the formula stated p.32. Toshow how, we first introduce a new event B = ∪
ni=1Ai.Now with the expression
for B inserted in (1) we get
P(
∪n+1
i=1 Ai
)
= P (B ∪ An+1) .
With exclusion-inclusion for two events (Page 22) we get
P (B ∪ An+1) = P (B) + P (An+1) − P (B ∩ An+1) . (2)
As
B ∩ An+1 = (∪ni=1Ai) ∩ An+1 = ∪
ni=1 (Ai ∩ An+1)
we have the formula stated at the top of Page 32. We now return our attentionto Equation (2). Since the exclusion-inclusion formula is assumed valid for n
events we can use this formula for the first term, thus for this term we get
P (B) = P (∪ni=1Ai) =
∑
i
P (Ai) −∑
i<j
P (Ai ∩ Aj) +∑
i<j<k
P (Ai ∩ Aj ∩ Ak) −
· · ·+ (−1)n+1P (A1 ∩ A2 . . . ∩ An) .
For the third term on the right hand side of Equation (2) we introduce theevents Bi = Ai ∩ An+1 such that this term can be expressed as
P (B ∩ An+1) = P (∪ni=1 (Ai ∩ An+1)) = P (∪n
i=1Bi) ,
implying that we can use the inclusion-exclusion formula for this term too.The proof is completed by writing down the expansion explicitly.
1
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