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Suppose we throw dice with six faces. Calculate these probabilities based on that information:

a. Six $2$’s or six $3$’s show up
b. Exactly five of any number show up
c. The numbers $1,2,3,4,5,6$ show up
d. Exactly three $1$’s shows up

Please correct me.

Sample space $6^6 =u$

a. $2/u$

b. $6^3/u$

c. $C(6,6)/u$

d. $C(6-1,3)/u$

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2 Answers

up vote 1 down vote accepted

The first answer is fine.

For the second problem, there are $6$ ways to choose the popular number, and for each of these ways there are $\binom{6}{5}$ ways to choose their locations. And for each of these there are $5$ ways to choose the unpopular number. So the numerator should be $180$.

For the third, there are $6!$ elements of our sample space that work, since the numbers can appear in any order.

For the fourth, the locations of the $1$'s can be chosen in $\binom{6}{3}$ ways. The other places can be filled in $5^3$ ways. So the numerator should be $\binom{6}{3}5^3$.

In all cases, as you indicated, the denominator is $6^6$.

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Your answer for (a) is correct. For (b) there are $6$ ways to choose which number shows up $5$ times, $5$ ways to choose the other number, and $6$ ways to choose which die shows the ‘wrong’ number, so this outcome can occur in $6\cdot5\cdot6=5\cdot6^2$ ways, and its probability is $\dfrac{5\cdot6^2}{6^6}=\dfrac5{6^4}$.

For (c) the six numbers can show up in any order, and there are $6!$ possible orders, so the correct answer is $\dfrac{6!}{6^6}$.

For (d) there are $\binom63$ ways to choose which $3$ of the dice show a $3$, and $5^3$ ways for the other $3$ dice not to show a $3$, so there are $5^3\dbinom63$ ways for this outcome to occur, and its probability is $\dfrac{5^3\dbinom63}{6^6}$.

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