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A pair of fair dice is tossed. Find the probability that the maximum of the two numbers is greater than $4$.

My attempted solution:

$E=\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5),$
$(1,6), (2,6), (3,6), (4,6), (5,6), (6,6),$
$(5,1), (5,2), (5,3), (5,4), (5,6),$
$(6,1), (6,2), (6,3), (6,4), (6,5)\}$

Here $n_s = 22.$

So, $P =\frac{22}{36} = \frac{11}{18}.$

But, the correct answer is, $\frac{5}{9}$.

What am I missing?

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    $\begingroup$ You counted some outcomes twice. E.g. (6, 5) $\endgroup$
    – Nighty
    Jun 8, 2018 at 9:47
  • $\begingroup$ One good trick to avoid double counting is to first list the good pairs $(i,j)$ with $i≤j$ . If $j=5$ there are five such, if $j=6$ there are six such. Of these exactly two have $i=j$ and we have to count the others twice (by symmetry). Thus there are $2+2\times 9=2+18=20$ good outcomes. $\endgroup$
    – lulu
    Jun 8, 2018 at 9:53
  • $\begingroup$ General note: in this case, it is certainly true that it is easiest to work with the complement. Still, it often happens that you have to count directly and in those cases it is good to learn methods of enumeration that reduce the chance of error (eliminating the chance of error is, alas, too much to hope for). $\endgroup$
    – lulu
    Jun 8, 2018 at 9:57
  • $\begingroup$ @lulu, in this case, it is certainly true that it is easiest to work with the complement. - why is that? $\endgroup$
    – user366312
    Jun 8, 2018 at 9:58
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    $\begingroup$ Note, for instance, that the complement method extends effortlessly to any number of dice. Had you asked about $10,000$ dice, for a random example, the answer would be $1-\left(\frac 23\right)^{10,000}$. Direct counting gets harder and harder with more dice. $\endgroup$
    – lulu
    Jun 8, 2018 at 10:02

2 Answers 2

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For it to be false, both dies have to be $1,2,3,4$. $$ p = \frac{4}{6} \times \frac{4}{6} = \frac{16}{36} = \frac{4}{9}, $$ therefore the probability of the statement being true is $$ 1 - \frac{4}{9} = \frac{5}{9}. $$

To calculate without using the complement, if the dies are $X$ and $Y$, it is $$ P(X > 4) + P(Y > 4) - P(X>4,\ Y>4) = \frac{1}{3} + \frac{1}{3} - \frac{1}{9} = \frac{6}{9} - \frac{1}{9} = \frac{5}{9}. $$ It is slightly trickier, if you add the two probabilities together, you have to knock off the area where they are both true. You can draw a Venn diagram to see that $$\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$ is the probability that both dice show greater than $4$.

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  • $\begingroup$ @yahoo.com No one said it cannot be solved without that. $\endgroup$
    – drhab
    Jun 8, 2018 at 9:58
  • $\begingroup$ it can be, if the dice are X and Y - it is P(X > 4) + P(Y > 4) - P(X>4 AND Y>4) = (1/3) + (1/3) - (1/9) = (6/9) - (1/9) = (5/9) It's slightly trickier, if you add the two probabvilites together, you have to knock off the area where they are both true. You can draw a Venn diagram to see this $\endgroup$
    – Cato
    Jun 8, 2018 at 10:00
  • $\begingroup$ your calculation above gives the probability that both dice are greater than 4 $\endgroup$
    – Cato
    Jun 8, 2018 at 10:01
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You double counted (6,5) and (5,6). It might be easier to argue about the complementary outcome.

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    $\begingroup$ Double counted (6,5) and (5,6). So it is $n_s=20$ and thus $P=\frac{20}{36}=\frac{5}{9}$. $\endgroup$
    – Thern
    Jun 8, 2018 at 9:51
  • $\begingroup$ @yahoo.com Find that (which one is easyer?) out yourself by comparing both methods and thinking it over. Then determine your own taste in this. $\endgroup$
    – drhab
    Jun 8, 2018 at 10:01

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