+ The sample space is every value of the die (1-6) paired with heads and paired with tails. That is, ${1,2,3,4,5,6} times {H,T}$, with cardinality 12.
+ There are $12^10$ outcomes.
+ If no participants roll a 5, then we omit any outcome in our sample space where the die outcome is 5, leaving us with 10 outcomes of the die and coin. Now we have $10^10$ outcomes. If at least 1 person rolls a 5, then we note that this is simply the complement of the previous result. So we have $12^10 - 10^10$ outcomes total.
]
+ #[
#set enum(numbering: "a)", spacing: 2em)
+ #[
The sample space can be represented as a 6-tuple where the position 1-6
represents balls numbered 1-6, and the value represents the square it
was sent to. So it's
$ {{x_1,x_2,x_3,x_4,x_5,x_6} : x_i in {1,2,3,4}}, i = 1,...6 $
]
+ #[
When the balls are indistinguishable, we can instead represent it as
4-tuples where the position represents the 1st, 2nd, 3rd, or 4th square,
and the value represents how many balls landed. Additionally the sum of all
2. Probability that the 4 numbers are in strictly increasing order.
To be strictly increasing, one simply chooses which 4 distinct numbers (out of 10) and then there is exactly _one_ way to list them in increasing order. Hence the favorable cases are #mitex(`\(\binom{10}{4}\)`). So
First find all 4‐element _subsets_ of #mitex(`\(\{1,\dots,10\}\)`) summing to 13:
#mitex(`\[
(1,2,3,7),\quad (1,2,4,6),\quad (1,3,4,5).
\]`)
There are exactly 3 such sets. Each set of 4 distinct numbers can appear in \(4!\) different orders among the draws. Thus the number of favorable ordered draws is \(3\times 4!=72.\) Therefore
2. Each player’s entire 13‐card hand is “all one suit.”
Since each suit has exactly 13 cards, this can only happen if one suit goes entirely to Player1, another suit to Player2, etc. There are 4 suits and 4 players, so the number of ways is simply the number of ways to _assign_ each suit to a distinct player:
#mitex(`\[
\text{Ways} \;=\;4!\;=\;24.
\]`)
3. Players1 and2 together get all the hearts.
There are 13 hearts and 39 other cards. Players3 and4 must then share the 39 non‐hearts only, while the 13 hearts + 13 of the non‐hearts go to Players1 and2. One convenient count is:
- Choose which 26 of the 39 non‐hearts go to Players3+4, then choose 13 of those for Player3 (and 13 for Player4).
- The remaining 13 non‐hearts plus the 13 hearts go to Players1+2, and we then choose which 13 go to Player1.
In binomial‐coefficient form:
#mitex(`\[
\text{Ways}
\;=\;
\binom{39}{13}\,\binom{26}{13}\,\binom{26}{13}.
\]`)
]
+ #[
#set enum(numbering: "a)", spacing: 2em)
Forming 10‐letter “words” from the letters \(\{B,A,C,O,N,R,U,L,E,S\}\).
1. Number of 10‐letter arrangements.
All 10 letters are distinct, so there are
#mitex(`\[
10!\;=\;3{,}628{,}800
\]`)
possible orderings.
2. Probability that the block “BACON” appears consecutively in that order.
Treat the five letters *BACON* as a single block plus the other 5
letters #mitex(`\(\{R,U,L,E,S\}\)`). That gives #mitex(`\(6\)`) total “items” to permute, so
#mitex(`\(6!\)`) orderings. There is only 1 way to arrange the block “BACON”
internally (since we want that exact order). Hence the favorable count is
#mitex(`\(6!=720\)`). Therefore
#mitex(`\[
P(\text{“BACON” together})
\;=\;\frac{6!}{10!}
\;=\;\frac{720}{3{,}628{,}800}
\;=\;\frac{1}{5040}.
\]`)]
+ #[
#set enum(numbering: "a)", spacing: 2em)
A succinct way to see the solution is to note that the six probabilities
#mitex(`\[
p_0,\,p_1,\,p_2,\,p_3,\,p_4,\,p_5
\]`)
form an arithmetic (nonincreasing) sequence, so one can write
#mitex(`\[
p_n \;=\;p_0 - n\,d\quad\text{for }n=0,1,2,3,4,5,
\]`)
where #mitex(`\(d \ge 0.\)`) The conditions then translate into the two equations
