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Consider this probability problem: An urn with 100 marbles. 90 are white, 10 are colored. Of the colored marbles three are red and seven are blue. When drawing out a single marble,

  • P(Red) = .03),
  • P(colored marble) = .1),
  • P(Red|a colored marble) = .3

Now the events, "colored marble" and "Red Marble given a colored marble" are clearly not independent, but if you multiply P(colored) and P(Red| colored marble) you get P(Red). This seems to be a broad, frequently occurring exception to the rule that one can only multiply the probability of independent events. What am I missing?

Tupelo Thistlehead
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    I think you are onfusing your rules. Independence says that $P(A\cap B)=P(A)\times P(B)$. Here, of course, one event is contained in the other, so the intersection is just "Red", hence, $P(\text {Red}\cap \text {coloured})=P(\text {Red})=.03$. – lulu May 12 '21 at 16:21

1 Answers1

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Given arbitrary events $A,B$ the equality $$ P(A\cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A) $$ is always true, so there is nothing wrong with your example.

However, for independent events we get $P(A|B) = P(A)$ and $P(B|A) = P(A)$ and the above formula simplifies to $$P(A\cap B) = P(A) \cdot P(B)$$

LegNaiB
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