∪1∞EiF=∪1∞EiFand∩1∞Ei∪F=∩1∞Ei∪F

∪1∞EiF=∪1∞EiF∩1∞Ei∪F=∩1∞Ei∪F

Q. 2.4 - A First Course in Probability

Q. 2.4

Question

$\cup_{1}^{\infty} E_{i} F = (\cup_{1}^{\infty} E_{i}) F$ and $\cap_{1}^{\infty} (E_{i} \cup F) = (\cap_{1}^{\infty} E_{i}) \cup F$

Step-by-Step Solution

Verified

Answer

$\cup_{1}^{\infty} E_{i} F = (\cup_{1}^{\infty} E_{i}) F \cap_{1}^{\infty} (E_{i} \cup F) = (\cap_{1}^{\infty} E_{i}) \cup F$

1Step 1 Given Information.

$\cup_{1}^{\infty} E_{i} F = (\cup_{1}^{\infty} E_{i}) F$ and $\cap_{1}^{\infty} (E_{i} \cup F) = (\cap_{1}^{\infty} E_{i}) \cup F$

2Step 2 Explanation.

first part

let $x \in (\cup_{1}^{\infty} E_{i}) F$ . Then $x \in F$ and $x \in \cup_{1}^{\infty} E_{i} x \in \cup_{1}^{\infty} E_{i}$ implies, $x \in E_{i}$ for some $i = j$ . This implies that $x \in E_{j} F$ . Thus $x \in E_{i} F$ for some $i$ and hence $x \in \cup_{1}^{\infty} E_{i} F$ . Therefore, $(\cup_{1}^{\infty} E_{i}) F \subseteq \cup_{1}^{\infty} E_{i} F$

conversely,

let $x \in \cup_{1}^{\infty} E_{i} F$ . Then $x \in E_{i} F$ for some $i = j$ . This implies that $x \in F$ and $x \in E_{j}$ .

i.e. $x \in F$ and $x \in E_{i}$ for somewidth="6" style="max-width: none;" $i$ . i.e. $x \in F$ and $x \in \cup_{1}^{\infty} E_{i}$ .

Hence, $x \in$ $(\cup_{1}^{\infty} E_{i}) F$

Therefore, from the above two arguments, $\cup_{1}^{\infty} E_{i} F = (\cup_{1}^{\infty} E_{i}) F$

3Step 3 Explanation.

second part

let $x \in (\cap_{1}^{\infty} E_{i}) \cup F$ . This means $x \in F$ or $x \in \cap_{1}^{\infty} E_{i}$ . If $x \in F$ then $x \in F \cup E_{i}$ for all $i$ and hence $x \in \cap_{1}^{\infty} (E_{i} \cup F)$ . If $x \in E_{i}$ for all $i$ then again, $x \in F \cup E_{i}$ for all $i$ and hence $x \in \cap_{1}^{\infty} (E_{i} \cup F)$ . Therefore $(\cap_{1}^{\infty} E_{i}) \cup F \subseteq \cap_{1}^{\infty} (E_{i} \cup F)$ .

conversely,

say $x \in \cap_{1}^{\infty} (E_{i} \cup F)$ . Then $x \in F \cup E_{i}$ for all $i$ . If $x \in F$ then clearly $x \in (\cap_{1}^{\infty} E_{i}) \cup F$ . If not, then since $x \in F \cup E_{i}$ all $i$ , $x$ has to be in $E_{i}$ for all $i$ . i.e. $x \in E_{i}$ for all $i$ and hence $x \in \cap_{1}^{\infty} E_{i}$ . Therefore, $x \in (\cap_{1}^{\infty} E_{i}) \cup F$ and $\cap_{1}^{\infty} (E_{i} \cup F) \subseteq (\cap_{1}^{\infty} E_{i}) \cup F$ .

from the above two arguments, $\cap_{1}^{\infty} (E_{i} \cup F) = (\cap_{1}^{\infty} E_{i}) \cup F$

Q. 2.3

Q. 2.5

Other exercises in this chapter

Q. 2.2

Prove the following relations:IfE⊂F, then Fc ⊂Ec

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Q. 2.3

F = FE ∪ FEcandE ∪ F = E ∪ EcF

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Q. 2.5

For any sequence of events E1, E2, ...,define a new sequence F1, F2, ... of disjoint events (that is, events such that FiFj&

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Q. 2.7

Use Venn diagrams(a) to simplify the expressions (E ∪ F)(E ∪ Fc);(b) to prove DeMorgan’s laws for eventsE a

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