# Thinking intuitively for probability multiplication rule

Suppose I have 2 dices, a Blue die and a Red die and I put them into the following events:

• Event A: {Blue Die rolled 1 or 2 or 3}
• Event B: {Red Die rolled 1 or 2 or 3}
• Event C: {Blue and Red dices rolled 1 or 2 or 3}

I know that the $P(A)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$ and $P(B)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$.

For the probability of event $C$, however, before I know if the events $A$ and $B$ are independent, I want to avoid using the $P(C)=P(A)\cdot P(B)$ formula and think of it intuitively.

I began my attempt to think of the problem 'intuitively' this way:

There are a total of $36$ outcomes for the 2 dices. The probability to get both red and blue dice rolled with value $1$, the probability is only $\frac { 1 }{ 36 }$. So, if I want both red and blue dice to be $1$ or $2$ or $3$, I use the addition rule: $\frac { 1 }{ 36 } +\frac { 1 }{ 36 } +\frac { 1 }{ 36 } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 }$ . So it seems like there is only 3 possibilities out of the 36 combinations of outcomes from the 2 dices and so $P(C)=\frac { 1 }{ 12 }$.

But if following the probability formula, event $C$ wants both Red and Blue dices to be 1 or 2 or 3, then $P(C)=P(A\cap B)=\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 4 }$, which is not $\frac { 1 }{ 12 }$!

What is wrong with my 'intuitive' way of thought?

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In your 'intuitive' method you have added up the possibilities of the dice showing (1,1), (2,2) or (3,3) but you haven't included possibilities like (1,2) or (3,2).

To count the possibilities correctly, you could organise your calculation like this:

• The red dice could show a 1, and then the blue dice could show 1, 2 or 3 (3 possibilities)
• The red dice could show a 2, and then the blue dice could show 1, 2 or 3 (3 possibilities)
• The red dice could show a 3, and then the blue dice could show 1, 2 or 3 (3 possibilities)

Adding these up, there are 9 possibilities out of a total of 36, and 9 / 36 = 1/4.

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$$P(A\textrm{ and }B) = P(A)P(B)$$
But since event $C$ wanted only both dices to be (1,1) or (2,2) or (3,3), why do I have to include other possibilities other than the 3 possibilities? – xenon Feb 2 '12 at 8:13
oh Thanks!! Does this mean that I will not be able to derive the probability of $C$ as $\frac{1}{12}$ mathematically from $P(A)$ and $P(B)$, or could I? – xenon Feb 2 '12 at 8:35