Q. 2.14

Question

Prove Proposition $4.4$ by mathematical induction.

Step-by-Step Solution

Verified

Answer

To prove the statement for $k + 1,$ considering first $k$ events as one, use Proposition $4.3 .$ , and then the presumption for $k$ twice.

1Step 1 Given Information.

Given that Proposition $4.4$ .

2Step 2 explanation.

We will apply mathematical induction to the number of sets. For $n = 2$ , see Proposition $4.3 .$ Now assume for $n = k$ we have. $P (E_{1} \cup E_{2} \cup \dots \cup E_{k}) = \sum_{i = 1}^{k} P (E_{i}) - \sum_{i_{1} < i_{2}} P (E_{i_{1}} E_{i_{2}}) + \dots + (- 1)^{r + 1} \sum_{i_{1} < \dots < i_{r}} P (E_{i_{1}} \dots E_{i_{r}}) + \dots + (- 1)^{k + 1} P (E_{1} \dots E_{k})$

For $n = k + 1 .$ Set $E = E_{1} \cup \dots \cup E_{k}$ and apply Proposition $4.3 .$ Then we get

$P (E_{1} \cup E_{2} \cup \dots \cup E_{k} \cup E_{k + 1}) = P (E) + P (E_{k + 1}) - P (E E_{k + 1})$

Using the induction hypothesis we get

$P (E_{1} \cup E_{2} \cup \dots \cup E_{k + 1}) = P (E_{k + 1}) + \sum_{i = 1}^{k} P (E_{i}) - \sum_{i_{1} < i_{2}} P (E_{i_{1}} E_{i_{2}}) + \dots + (- 1)^{r + 1} \sum_{i_{1} < \dots < i_{r}} P (E_{i_{1}} \dots E_{i_{r}}) + \dots + (- 1)^{k + 1} P (E_{1} \dots E_{k}) - P (E E_{k + 1})$

Now observe that

$P (E E_{k + 1}) = P ((E_{1} \cup \dots \cup E_{k}) E_{k + 1}) = P (E_{1} E_{k + 1} \cup \dots \cup E_{k} E_{k + 1}) = \sum_{i = 1}^{k} P (E_{i} E_{k + 1}) - \sum_{i_{1} < i_{2}} P (E_{i_{1}} E_{i_{2}} E_{k + 1}) + \dots + (- 1)^{r + 1} \sum_{i_{1} < \dots < i_{r}} P (E_{i_{1}} \dots E_{i_{r}} E_{k + 1}) + \dots + (- 1)^{k + 1} P (E_{1} \dots E_{k} E_{k + 1})$

Combining this with the identity above we get

$P (E_{1} \cup E_{2} \cup \dots \cup E_{k + 1}) = \sum_{i = 1}^{k + 1} P (E_{i}) - \sum_{i_{1} < i_{2}} P (E_{i_{1}} E_{i_{2}}) + \dots + (- 1)^{r + 1} \sum_{i_{1} < \dots < i_{r}} P (E_{i_{1}} \dots E_{i_{r}}) + \dots + (- 1)^{k + 2} P (E_{1} \dots E_{k + 1})$

Hence we get the desired identity by mathematical induction.

Q. 2.9

Q. 2.11

Other exercises in this chapter

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Q. 2.9

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Q. 2.11

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Q. 2.12

Show that the probability that exactly one of the events E or F occurs equals P(E) + P(F) − 2P(EF).

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