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According to my textbook,

Two events are independent if the occurrence of one does not affect the probability of the other.

Or we could also say,

Two events are not independent if the occurrence of one does affect the probability of the other.

Suppose we draw two cards in succession from a deck (without replacement). Let

  • E = "the first draw is a Jack"
  • F = "the second draw is a Jack"

We can conclude that E and F are not independent. Strangely, my argument and the textbook's argument for the same conclusion are different. I know my reasoning is incorrect somehow but I'm not sure why.

My Argument: I simply assume E occurs and assume E does not occur to see whether this affects the probability of F.

  • if E occurs then P(F)=3/51
  • if E does not occur then P(F)=4/51

Notice the "occurrence of one does affect the probability of the other". That is, the occurrence of E does affect the probability of F. Thus, I can then safely conclude, that the events are not independent based on the definition above, cased closed.

Textbook Argument: Notice that P(E)=4/52 and P(F|E)=3/51. Since they are not equal, the events are not independent. I don't understand why this proves it. They aren't even the same draw. The former is the first draw and the latter is the second draw. I don't see why we should be comparing them at all. It doesn't seem to be applicable to the definition above.

I must be assuming something, misunderstanding the definition, or missing something obvious. Can someone help me sort this out?

B flat
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2 Answers2

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Your reasoning is correct.

The textbook's reasoning is not, but that may be due to a misprint or a misreading. It would be correct if it said:

Notice that $P(F)=\frac{4}{52}$ and $P(F|E)=\frac{3}{51}$. Since they are not equal, the events are not independent.

Note that the first is $P(F)$, i.e. the probability of the second draw being a jack, without any knowledge of what the first draw is. If you know that the first draw was a jack, this changes the probability to $P(F|E)=\frac{3}{51}$. The probability has changed value, so the events are not independent.

Note that knowing that the first draw is NOT a jack also changes the probability, namely to $P(F|\overline E)=\frac{4}{51}$. This again shows the two events are dependent.

We have three probabilities: $P(F)$, $P(F|E)$, and $P(F|\overline E)$. If $E$ and $F$ are independent these would all be equal, and if $E$ and $F$ are dependent they would all be different. You argued that the latter two differed, while the book (after the minor correction) argued that the first two differed. Both approaches are valid.

Jaap Scherphuis
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The sentences you've quoted are intuitive descriptions of what independence means, but they are not mathematical definitions.

The mathematical definition of two events $A$ and $B$ being independent is any one of the equivalent equalities $$ P\bigl(A\cup B\bigr)=P(A)P(B) \quad\text{or}\quad P\bigl(A\mid B\bigr) = P(A) \quad\text{or}\quad P\bigl(B\mid A\bigr)=P(B). $$

Greg Martin
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  • Thank you. Yes, I know the formulas for the solid definition. However, I think it's worthwhile finding an equivalent expression in words something like the quotes above in the textbook It seems like there is either an assumption in that quote or they are missing something that can be added to make it equivalent to the solid definition. Maybe not, but it just seems like it would be possible. – B flat Mar 17 '22 at 07:41