# What is incorrect with my attempt to calculate $Pr(A \cap B)$ without any formulae?

Suppose I have 1 Blue and 1 Red die, which I put 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: {Red and blue die are equal, and both show 1, 2 or 3}

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

I wish to calculate $P(C)$ intuitively, without relying on the formula $P(C)=P(A)\cdot P(B)$.

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 $3$, it seems like there is only 3 possibilities out of the 36 combinations of outcomes from the 2 dices. So I use the addition rule: $Pr(C) = \frac { 1 }{ 36 } +\frac { 1 }{ 36 } +\frac { 1 }{ 36 } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 }$.

But if I use the probability formula, event $C$ needs 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' solution?

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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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Your 'non-intuitive' calculation is incorrect. When you use the rule

$$P(A\textrm{ and }B) = P(A)P(B)$$

then the left-hand side represents the possibilities that A and B both occur.
Since your event A is "The blue dice is 1, 2 or 3" and event B is "The red dice is 1, 2 or 3"
then P(A)P(B) is the event "The blue dice is 1, 2 or 3 AND the red dice is 1, 2 or 3"
which allows possibilities like "Blue = 1, Red = 2" which don't form part of your event C, which is "The red and blue dice are equal, and both show 1, 2 or 3".

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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
Ah, okay. You haven't been clear enough in your description of event C - it currently reads like "C = (Blue dice is 1, 2 or 3) AND (Red dice is 1, 2 or 3)" whereas what you meant is "C = (Blue and red are 1) OR (Blue and red are 2) OR (Blue and red are 3)". I will edit my answer. – Chris Taylor Feb 2 '12 at 8:17
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