The sample space of the experiment of throwing a green and a red dice has 36 elements. The event, say $A$, that the sum $x+y>8$ in $(x,y)$, where $x$ is an outcome of the green die and $y$ that of red die will occur has the probability:


And the probability of the event, say C, that a number greater than 4 will turn up on the green die is:

$$P(C)=\frac{12}{36} = \frac{1}{3}$$

And the probability of the intersection $A \cap C$ is:

$$P(A \cap C) = \frac{7}{36}$$

which is not equal to the product of the probabilities of $A$ and $C$, that is

$$P(A) \cdot P(C)=\frac{5}{54} \neq \frac{7}{36}$$

Is the law of product even applicable here? If so, how?

  • 1
    $\begingroup$ The problem is that the events A and C that you selected are NOT independent, So the product rule may not apply. $\endgroup$ – Michael Chernick Mar 23 '18 at 17:01

The events A & C are dependent event. That is, if one of them occurs first, the odds of second event are changed. The law of product for dependent events is:

P(A∩C) = P(C).P(A|C)

Where P(A|C) is conditional probability of the event A given event C has already occurred.

If event C (x > 4) has already occurred, then we have a sample space of 12 instead of 36. This is already clear in your calculation of P(C).

There are 7 Dice throw cases where event A (x + y > 8) occurs, given that event C (x > 4) already occurred: (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)

hence P(A|C) = 7/12

P(A∩C) = P(C).P(A|C) = 1/3 . 7/12 = 7/36

which is same as the P(A∩C) calculated independently.

By the way, we can also prove the same other way around, that is: P(A∩C) = P(A).P(C|A) (But I chose the case which I was comfortable with.)

PS: pls disregard my clumsy writing, i still have not adapted to write fractions & other mathematic symbols in this forum.


Product of probabilities equals their joint probability only for independent events. In fact, it is a part of definition of independence:

Two events A and B are independent (often written as $A \perp B$ or $A \perp\!\!\!\perp B$) if their joint probability equals the product of their probabilities:

$$\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$$

  • $\begingroup$ Could you explain why $A$ and $C$ are not independent events? $\endgroup$ – Samama Fahim Mar 23 '18 at 17:44
  • $\begingroup$ @SamamaFahim because the first event "tells" you something about possible outcomes of the first one. Imagine more extreme example A=z<5 and B=z<4.9999..., if I told you that A is true, would it change anything about your bet on B? $\endgroup$ – Tim Mar 23 '18 at 17:57

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