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A die is rolled. What is the probability that the number is even and less than 4?

Event $A$: Numbers on a die that are even: 2, 4, 6
Event $B$: Numbers on a die that are less than 4: 1, 2, 3
There is only one number (2) that is in both events A and B.
Total outcomes $S$: Numbers on a die: 1, 2, 3, 4, 5, 6 (total = 6)

Ok, so obviously the one possible outcome in this example is $1/6$. But if I use the rule of multiplication which states that if the sample space is the same, which I think it is, then $P(a)P(b)$ should give me the right answer, which it doesn't: $1/4$.

(My logic here is that less than 4 is 50% chance and the even numbers are 50% chance).

So the big question is, why doesn't the multiplication rule work here?

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    $\begingroup$ The answer to your question is that the events even and less than $4$ are not independent. $\endgroup$ – André Nicolas Jul 20 '15 at 20:54
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    $\begingroup$ The simple rules only work if the events are independent. Say Event X was "the number you throw is less than 3" and Event Y was "the number you throw is greater than 3". Then the probability of both is 0. $\endgroup$ – lulu Jul 20 '15 at 20:54
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There is a way of recovering a version of the multiplication rule.

Let $A$ and $B$ be as in your post. We cannot write $\Pr(A\cap B)=\Pr(A)\Pr(B)$, since $A$ and $B$ are not independent.

However, we have $$\Pr(A\cap B)=\Pr(A)\Pr(B|A),$$ where $\Pr(B|A)$ means the (conditional) probability $B$ happens given that $A$ happens.

Given that $A$ happened, meaning the number is even, the probability it is between $1$ and $3$ is $1/3$, since only one-third of evens are between $1$ and $3$. So $\Pr(A\cap B)=(1/2)(1/3)$.

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Observe that Event A and Event B are statistically independent iff $P(A \cap B) = P(A)P(B)$, which is decidedly not the case here ($P(A\cap B)=P(2)=1/6, P(A)P(B)=1/4$).

As a result you can not use the multiplication rule for independent events.

https://en.wikipedia.org/wiki/Independence_%28probability_theory%29#More_than_two_events

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Consider these two events:

  • $A$: the result of the die is not odd: $\{2, 4, 6\}, P(A)=1/2$
  • $B$: the result of the die is divisible by 2: $\{6, 4, 2\}, P(B)=1/2$

Now if we were able to use the multiplication rule we would say $\mathsf P(A\cap B) \mathop{=}^{?} \mathsf P(A)\,\mathsf P(B) = 1/4$.

This is, of course, nonsense.   The questioned equality does not hold.

The events are the same, so the probability of their intersection must be the probability of the events themselves: $\mathsf P(A\cap B) = 1/2$.

So clearly the multiplication rule is not useable when the events are not independent.

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To apply multiplication rule, the intersection of values of the individual sets should be null matrix. i.e A \cap B = null matrix

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