In chapter 13 (beginning of probability theory) of the book "challenge and thrill of pre college mathematics", after theorem 3 (i.e. for any 2 events E and F, P(E union F)=P(E)+P(F)+P(E intersection F)), there is an illustration in which "P(throwing either an even number or a sum of 2,3,4,with two dice)" has to be calculated.

The illustration is : E=event{2,4,6}, F=event{2,3, 4}. Then E U F={2,3,4,6},P(E U F) =11/36. P(E) =9/36, P(F) =6/36, P(EF) =P(2, 4) =4/36. P(E) +P(F) -P(EF) =(9+6-4) /36=11/36=P(E U F).

My questions are, 1.) what does it mean "to throw an even number on 2 dice"?(should the sum of 2 numbers on 2 dice be even?)

2.) "sum of 2,3,4" means all possible sums from 2,3,4(i.e. 2+3=5,3+3=6,3+4=7,4+4=8,3+3=2+3+4=9 etc.),or sum of numbers on the dice should be 2,3 or 4?


This is what your textbook means:

Let's throw 2 fair dice:

  • Event $E$: the sum of the "2 dice's face up" is even

  • Event $F$: the sum "2 dice's face up" is 2,3 or 4


$$\mathbb{P}[E \cup F]=\mathbb{P}[E ]+\mathbb{P}[ F]-\mathbb{P}[E \cap F]=\frac{1}{2}+\frac{6}{36}-\frac{4}{36}=\frac{5}{9}$$

  • 1
    $\begingroup$ Alas, that (sensible) reading does not appear to be compatible with the illustration. Writing $E=\{2,4,6\}$ does not make sense if you are requiring that the sum be even. It leaves off $\{8,10,12\}$. $\endgroup$ – lulu Mar 3 at 11:44
  • $\begingroup$ You are mistaking: $P(E)=\frac14\ne\frac12$. $\endgroup$ – user Mar 3 at 11:54
  • $\begingroup$ @user : sorry, If I did not read wrong, $E$ is the event: sum of 2 dice is even...thus $P(E)=0.5$ $\endgroup$ – tommik Mar 3 at 11:57
  • $\begingroup$ The author's solution (given in the comments to OP) claims $\mathbb P(E)=\frac9{36}$ . $\endgroup$ – user Mar 3 at 12:01

I think the authors are subtly trying to point out that the probability of throwing two dice both even is the same as throwing two dice that then sum to $2,4$ or $6$, which is easy to see by any mapping of:


onto $$11,13,22,31,15,24,33,42,51$$

such as $(x, y)\to (x+\min(x,y), y+\min(x,y))$.

  • $\begingroup$ This does not suit. In this case $\mathbb P(E\cap F)=\frac1{36}$ as the only intersection of two events is $(2,2)$. $\endgroup$ – user Mar 3 at 13:01
  • $\begingroup$ @user; they have that in the repeated line 4 - admittedly the 'U' is upside-down! $\endgroup$ – JMP Mar 3 at 13:08
  • $\begingroup$ I am quite sure the corresponding value is given in $P(EF)=P(2,4)=\frac4{36}$. Otherwise it would not be used in PIE. (Hier $(X,Y)$ means the value of the sum rather than the result of the throw as in my example). $\endgroup$ – user Mar 3 at 13:12

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