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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 1

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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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  • $\begingroup$ Ugh, thanks. I feel dumb now. I'll accept your answer in eight minutes when the site lets me. $\endgroup$
    – jsmith95
    Feb 9, 2012 at 4:47
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    $\begingroup$ @jsmith95: Don't feel dumb. It was just an oversight (: $\endgroup$
    – JavaMan
    Feb 9, 2012 at 4:48

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