A, B and C are three independent events such that, $P(A) = \frac12, P(B) = \frac13, P(C) = \frac17$

Find $P((A\cup B)\cap C)$.

My Answer:

$$P((A\cup B)\cap C) = P((A\cup B))\cdot P(C) = \frac23 \cdot \frac17 =\frac2{21}$$


$$P((A\cup B)\cap C) =P((A\cap C)\cup (B\cap C)) = \frac1{14} + \frac1{21} - \frac1{14}\cdot \frac1{21} = \frac{34}{14\cdot 21} $$

Why are the two methods giving different answers?


$A \cap C$ and $B \cap C$ are not independent.

We have $P(A \cap B \cap C)=\frac1{42}$.

Hence, the second computation should be $$\frac1{14}+\frac1{21}-\frac1{42}=\frac1{14}+\frac1{42}=\frac{4}{42}=\frac{2}{21}$$

  • $\begingroup$ Hi many thanks. Can you share some materials for this? How do we know that A intersection C and B intersection C are not independent? $\endgroup$
    – Dom Jo
    Mar 14 '20 at 9:02
  • $\begingroup$ Suppose they are independent, then your computation would reach a contradiction. I just work it out from the definition. $\endgroup$ Mar 14 '20 at 9:05
  • $\begingroup$ But still, can you share some links or materials? I still cannot wrap my head around this $\endgroup$
    – Dom Jo
    Mar 14 '20 at 9:06
  • $\begingroup$ Is it that P((A∩C) U (B∩C)) == P(A∩C) + P(B∩C) - P(A∩C∩B) ?? $\endgroup$
    – Dom Jo
    Mar 14 '20 at 9:08
  • $\begingroup$ yes, we use that formula that you just stated. not sure how helpful is this link. $\endgroup$ Mar 14 '20 at 9:14

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