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Problem:Let $E$ and $F$ be two events such that $P(E)=0.7$, $P(F)=0.4$ and $P(E∩F')=0.4$. Then, $P(F|E∪F')$ is equal to

  1. $1/2$
  2. $1/3$
  3. $1/4$
  4. $1/5$

My Attempt: We know that $$P(F|E∪F')=\frac{P(F\cap\{E\cup F'\})}{P(E\cup F')}.$$ Now $P(E\cup F')=P(E)+P(F')-P(E\cap F')=0.7+0.6-0.4=0.9$ and $P(F\cap\{E\cup F'\})=P(\{F\cap F'\}\cup E)=P(\phi\cup E)=P(E).$ And so $$P(F|E∪F')=\frac{P(F\cap\{E\cup F'\})}{P(E\cup F')}=\frac{P(E)}{0.9}=7/9.$$ But apparently this is not mentioned in any of the options. Where am I going wrong?

  • 0
    Your mistake is here: $$F \cap (E \cup F') = (F \cap F') \cup E$$ is not correct (how did you get that?). The correct expression is $$F \cap (E \cup F') = (F \cap E) \cup (F \cap F') = F \cap E$$. – 2017-02-02

2 Answers 2

1

Using distributive law -

$P(F\cap\{E\cup F'\} = P(F \cap E) \cup P(F \cap F')$

$P(F\cap\{E\cup F'\} = P(F \cap E)$

So your mistake here.

Solution -

$P(E \cap F') = P(E) - P(E \cap F)$

$0.4 = 0.7 - P(E \cap F)$

$P(E \cap F) = 0.3$

Now,

$P(F|E∪F')=\frac{P(F\cap\{E\cup F'\})}{P(E\cup F')}$

$=\frac{0.3}{0.9}=\frac 13$

1

Your error occurs in simplifying $P(F \cap \{ E\cup F' \})$. The distribution law states $F \cap \{ E\cup F' \} = \{F \cap E\} \cup \{F \cap F'\} = \{F \cap E\} \cup ϕ = \{F \cap E\}$.

Then $P(F \cap \{ E\cup F' \})= P(\{F \cap E\})=P(E)-P(\{F' \cap E\})=0.3$.

Giving a final answer of $\frac{1}{3}$.