Chapter 2: Q. 2.12 (page 53)

Show that the probability that exactly one of the events $E o r F$ occurs equals $P (E) + P (F) - 2 P (E F) .$

Short Answer

Expert verified

Note that the event in which only one $E$ , $F$ occurs is $P ((E \cup F) (E F)^{c})$

Use Axiom $3$ for $(E \cup F) (E F)^{c})$ & $E \cup F$ and Proposition $4.3$ .

Step by step solution

Given Information.

The probability that exactly one of the events $E o r F$ occurs equals $P (E) + P (F) - 2 P (E F) .$ .

Explanation.

For any events $E$ and $F$ :

$P ((E \cup F) (E F)^{c}) = P (E) + P (F) - 2 P (E F)$

Where the left-hand side is the event where precisely one event $E$ & $F$ occurs.

Because $(E \cup F) (E F)^{c}$ is an event that occurs if $E$ or $F$ occur and not both $E$ andlocalid="1649246913116" $F$ occur.

Firstly note that

$(E \cup F) (E F)^{c} \cap E F = \emptyset (E \cup F) (E F)^{c} \cup E F = E \cup F$

Therefore

$P (E \cup F) = P ((E \cup F) (E F)^{c}) + P (E F) P ((E \cup F) (E F)^{c}) = P (E \cup F) - P (E F) = P (E) + P (F) - P (E F) - P (E F) = P (E) + P (F) - 2 P (E F)$

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Show that the probability that exactly one of the events $E o r F$ occurs equals $P (E) + P (F) - 2 P (E F) .$

Short Answer

Step by step solution

Given Information.

Explanation.

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

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Step by step solution

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Show that the probability that exactly one of the events $E o r F$ occurs equals $P (E) + P (F) - 2 P (E F) .$