This is the method I thought about the problem. Allow $X$ it is in the random variable describing the variety of occurrences the a number $a$, where $1 \\leq a \\leq 6$. Then the probability the $a$ occurs double is the binomial

$P(X=2) = 5 \\choose 2(\\frac16)^2(\\frac56)^3 = 0.16$

Now there are 6 possibilities for $a$, therefore the probability of rojo a pair of any type of number between one and also six is $0.16*6 = 0.96$.

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I don\"t understand where i went wrong however the final answer I acquire seems wrong intuitively to me. I would certainly be thankful for some correction or verification.

You can\"t simply add the probabilities $P(X=2)$ together $a$ varieties over $1,\\dots, 6$, since the events of rolling 2 of a number aren\"t support exclusive (it is feasible to roll two ones

*and*two fives). Furthermore, you would want to look in ~ $P(X\\ge 2)$, no $P(X=2)$, due to the fact that having 3 or much more of a number also counts as having a matching pair.

I think the best method to do this trouble is very first find the probability there is *no* lifwynnfoundation.orgcing pair, climate subtract indigenous 1. Because that the dice come be every different, there are 6 options for the first die, 5 for the second, 4 for the third$\\dots$, and $6^5$ feasible rolls total, therefore the probability is$$1-\\frac6\\cdot5\\cdot4\\cdot3\\cdot26^5$$

The fastest and also safest an approach for solving this would certainly be Mike Earnest\"s solution or something favor it.

If you desire to add the probabilities of acquiring your pair rather than individually the probability that *not* gaining a pair, you have to avoid double-counting.As has already been mentioned, you deserve to roll two $1$s *and* two $2$s in the same five rolls.To compute the probability of precisely two $1$s and also *no other* pairs,you have $5$ options for the first die the is not in the pair, but for every of thosethere remain only $4$ options for the following die, and at the finish there areonly $3$ selections for the last die, so$$P(\\mboxpair the twos) = \\binom52 \\left(\\frac16\\right)^2 \\cdot \\frac56 \\cdot \\frac 46 \\cdot \\frac 36.$$This event is support exclusive from having a pair of $2$s, so currently it is yes sir tomultiply through $6$ to find the probability the having specifically one pair of precisely one number.

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If you want *at least* one pair, climate you must also add the probability ofhaving precisely one pair each of two different numbersand the probability of having actually three or an ext of a single number(which you might further subdivide into precisely three, specifically four, or fiveof the same number).This is much more tedious and also error-prone 보다 the method of individually theprobability of no pairs.