# You roll two dice. What is the probability of the event that they have no common factor greater than unity?

This question appears in Stirzaker's Elementary Probability (2nd Edition). It is worked example 1.8 (p40).

My solution and answer:

The answer is $\mathcal{P}(A^c)$, where:

$A = \lbrace (2, 4), (4,2), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4) \rbrace$.

There are 8 pairs, so the probability is $1 - 8/36 = 28/36 = 7/9$.

The textbook's solution and answer:

It is routine to list the outcomes that do have a common factor greater than unity. They are 13 in number, namely:

$\lbrace (i,i); i \geq 2 \rbrace, (2, 4), (4,2), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4)$.

This is the complementary event, so by (1.4.5) the required probability is

$1 - 13/36 = 23/36$

Where'd the number 13 come from? Stirzaker enumerates the same set, but says there are 13 members. Where does he get the extra 5?

Screenshot of solution:

https://imgur.com/QpvLA

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## 1 Answer

The extra five are given by $\{(2,2); (3,3); (4,4); (5,5); (6,6)\}$, which is precisely the set $\{(i,i), i \geq 2\}$.

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Ugh, thanks. I feel dumb now. I'll accept your answer in eight minutes when the site lets me. –  jsmith95 Feb 9 '12 at 4:47
@jsmith95: Don't feel dumb. It was just an oversight (: –  JavaMan Feb 9 '12 at 4:48