Let the sequence of events E1, E2, . . . , En be independent, and assume that P(Ei)=1i+1. Show that P(E1∪···∪En)=nn+1

Given: E1,E2…En, are independent events P(Ei)=1i+1 To proof: P(E1U…UEn)=nn+1 Use the Complement rule: P(Ac)=P(¬A)=1−P(A) P((Ei)c)=1−P(Ei)=1−(1i+1)=i+1i+1−(1i+1)=ii+1 Since the events E1,E2…,En are independent, the events (E1)c,(E2)c…(En)c are alslo independent. We can use the multification rule for independent events P(A∪B)=P(AandB)=P(A)xP(B) P((E1)c)∩…∩((En)c)=P(E1)c⋅…⋅P((En)c)=11+1⋅(22+1)⋅(33+1)⋅…⋅(nn+1=12⋅23⋅34⋅…⋅nn+1=1(2)(3)(4)…(n)2(3)(4)…(n+1)=1n+1 Use the complement rule: P(Ac)=P(¬A)=1−P(A) P(E1…∪En)=P(((E1)c)∩…∩(En)c))=1−P(((E1)c)∩…∩(En)c)=1−(1n+1)=n(n+1)

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Kyran Hudson

Answered question

2021-02-12

Let the sequence of events E1, E2, . . . , En be independent, and assume that

P (E i) = \frac{1}{i + 1}

. Show that

P (E 1 \cup \cdot \cdot \cdot \cup E n) = \frac{n}{n + 1}

Answer & Explanation

grbavit

Skilled2021-02-13Added 109 answers

Given: $E_{1}, E_{2} \dots E n,$ are independent events $P (E i) = \frac{1}{i + 1}$
To proof: $P (E 1 U \dots U E n) = \frac{n}{n + 1}$
Use the Complement rule: $P (A^{c}) = P (\neg A) = 1 - P (A)$
$P ({(E i)}^{c}) = 1 - P (E i) = 1 - (\frac{1}{i + 1}) = \frac{i + 1}{i + 1} - (\frac{1}{i + 1}) = \frac{i}{i + 1}$
Since the events $E_{1}, E_{2} \dots, E_{n}$ are independent, the events ${(E_{1})}^{c}, {(E_{2})}^{c} \dots {(E_{n})}^{c}$ are alslo independent.
We can use the multification rule for independent events $P (A \cup B) = P (A and B) = P (A) x P (B)$
$P ({(E 1)}^{c}) \cap \dots \cap ({(E n)}^{c}) = P {(E 1)}^{c} \cdot \dots \cdot P ({(E n)}^{c}) = \frac{1}{1 + 1} \cdot (\frac{2}{2 + 1}) \cdot (\frac{3}{3 + 1}) \cdot \dots \cdot (\frac{n}{n + 1} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \dots \cdot \frac{n}{n + 1} = \frac{1 (2) (3) (4) \dots (n)}{2 (3) (4) \dots (n + 1)} = \frac{1}{n + 1}$
Use the complement rule: $P (A^{c}) = P (\neg A) = 1 - P (A)$
$P (E_{1} \dots \cup E_{n}) = P (((E_{1})^{c}) \cap \dots \cap (E_{n})^{c})) = 1 - P (((E_{1})^{c}) \cap \dots \cap (E_{n})^{c}) = 1 - (\frac{1}{n + 1}) = \frac{n}{(n + 1)}$

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