12. Probability and counting

Answer: $1680$

The first place has $8$ candidates, the next $7$, and so on for $4$ places: $\dfrac{8!}{(8 - 4)!} = 1680$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $70$
• Arranged all of them rather than only the chosen ones → $40320$

Answer: $70$

A committee is a SET, so order does not matter: $\binom{8}{4} = \dfrac{8!}{4!\,(8 - 4)!} = 70$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $1680$
• Multiplied the two numbers → $32$

Answer: $\frac{1}{9}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is exactly 5 gives $4$, so the probability is $\dfrac{4}{36} = \frac{1}{9}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $\frac{4}{11}$
• Used 12 outcomes, one per face of each die → $\frac{1}{3}$

Answer: $\frac{3}{8}$

Each of the $2^{3} = 8$ sequences is equally likely, and $\binom{3}{1} = 3$ of them have exactly $1$ heads. So the probability is $\frac{3}{8}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{1}{3}$
• Used $2n$ outcomes instead of $2^n$ → $\frac{1}{2}$

Answer: $336$

The first place has $8$ candidates, the next $7$, and so on for $3$ places: $\dfrac{8!}{(8 - 3)!} = 336$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $56$
• Arranged all of them rather than only the chosen ones → $40320$

Answer: $45$

A committee is a SET, so order does not matter: $\binom{10}{2} = \dfrac{10!}{2!\,(10 - 2)!} = 45$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $90$
• Multiplied the two numbers → $20$

Answer: $\frac{7}{12}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is at least 7 gives $21$, so the probability is $\dfrac{21}{36} = \frac{7}{12}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $1$
• Used 12 outcomes, one per face of each die → $\frac{7}{4}$

Answer: $\frac{5}{16}$

Each of the $2^{5} = 32$ sequences is equally likely, and $\binom{5}{3} = 10$ of them have exactly $3$ heads. So the probability is $\frac{5}{16}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{3}{5}$
• Used $2n$ outcomes instead of $2^n$ → $1$

Answer: $56$

The first place has $8$ candidates, the next $7$, and so on for $2$ places: $\dfrac{8!}{(8 - 2)!} = 56$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $28$
• Arranged all of them rather than only the chosen ones → $40320$

Answer: $15$

A committee is a SET, so order does not matter: $\binom{6}{2} = \dfrac{6!}{2!\,(6 - 2)!} = 15$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $30$
• Multiplied the two numbers → $12$

Answer: $\frac{5}{36}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is exactly 8 gives $5$, so the probability is $\dfrac{5}{36} = \frac{5}{36}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $\frac{5}{11}$
• Used 12 outcomes, one per face of each die → $\frac{5}{12}$

Answer: $\frac{5}{16}$

Each of the $2^{5} = 32$ sequences is equally likely, and $\binom{5}{2} = 10$ of them have exactly $2$ heads. So the probability is $\frac{5}{16}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{2}{5}$
• Used $2n$ outcomes instead of $2^n$ → $1$

Answer: $30$

The first place has $6$ candidates, the next $5$, and so on for $2$ places: $\dfrac{6!}{(6 - 2)!} = 30$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $15$
• Arranged all of them rather than only the chosen ones → $720$

Answer: $20$

A committee is a SET, so order does not matter: $\binom{6}{3} = \dfrac{6!}{3!\,(6 - 3)!} = 20$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $120$
• Multiplied the two numbers → $18$

Answer: $\frac{1}{12}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is exactly 10 gives $3$, so the probability is $\dfrac{3}{36} = \frac{1}{12}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $\frac{3}{11}$
• Used 12 outcomes, one per face of each die → $\frac{1}{4}$

Answer: $\frac{5}{16}$

Each of the $2^{6} = 64$ sequences is equally likely, and $\binom{6}{3} = 20$ of them have exactly $3$ heads. So the probability is $\frac{5}{16}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{1}{2}$
• Used $2n$ outcomes instead of $2^n$ → $\frac{5}{3}$

Answer: $210$

The first place has $7$ candidates, the next $6$, and so on for $3$ places: $\dfrac{7!}{(7 - 3)!} = 210$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $35$
• Arranged all of them rather than only the chosen ones → $5040$

Answer: $35$

A committee is a SET, so order does not matter: $\binom{7}{4} = \dfrac{7!}{4!\,(7 - 4)!} = 35$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $840$
• Multiplied the two numbers → $28$

Answer: $\frac{5}{12}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is at least 8 gives $15$, so the probability is $\dfrac{15}{36} = \frac{5}{12}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $1$
• Used 12 outcomes, one per face of each die → $\frac{5}{4}$

Answer: $\frac{15}{64}$

Each of the $2^{6} = 64$ sequences is equally likely, and $\binom{6}{2} = 15$ of them have exactly $2$ heads. So the probability is $\frac{15}{64}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{1}{3}$
• Used $2n$ outcomes instead of $2^n$ → $\frac{5}{4}$

Answer: $504$

The first place has $9$ candidates, the next $8$, and so on for $3$ places: $\dfrac{9!}{(9 - 3)!} = 504$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $84$
• Arranged all of them rather than only the chosen ones → $362880$

Answer: $126$

A committee is a SET, so order does not matter: $\binom{9}{4} = \dfrac{9!}{4!\,(9 - 4)!} = 126$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $3024$
• Multiplied the two numbers → $36$

Answer: $\frac{5}{18}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is at least 9 gives $10$, so the probability is $\dfrac{10}{36} = \frac{5}{18}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $\frac{10}{11}$
• Used 12 outcomes, one per face of each die → $\frac{5}{6}$

Answer: $\frac{3}{32}$

Each of the $2^{6} = 64$ sequences is equally likely, and $\binom{6}{1} = 6$ of them have exactly $1$ heads. So the probability is $\frac{3}{32}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{1}{6}$
• Used $2n$ outcomes instead of $2^n$ → $\frac{1}{2}$

Answer: $72$

The first place has $9$ candidates, the next $8$, and so on for $2$ places: $\dfrac{9!}{(9 - 2)!} = 72$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $36$
• Arranged all of them rather than only the chosen ones → $362880$

Answer: $120$

A committee is a SET, so order does not matter: $\binom{10}{3} = \dfrac{10!}{3!\,(10 - 3)!} = 120$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $720$
• Multiplied the two numbers → $30$

Answer: $\frac{5}{6}$

There are $6 \times 6 = 36$ equally likely outcomes. Counting the ones whose total is at least 5 gives $30$, so the probability is $\dfrac{30}{36} = \frac{5}{6}$.

Common mistakes, and the answer each one gives:

• Counted the 11 possible TOTALS as equally likely → $1$
• Used 12 outcomes, one per face of each die → $\frac{5}{2}$

Answer: $\frac{5}{32}$

Each of the $2^{5} = 32$ sequences is equally likely, and $\binom{5}{1} = 5$ of them have exactly $1$ heads. So the probability is $\frac{5}{32}$.

Common mistakes, and the answer each one gives:

• Divided the count of heads by the number of tosses → $\frac{1}{5}$
• Used $2n$ outcomes instead of $2^n$ → $\frac{1}{2}$

Answer: $42$

The first place has $7$ candidates, the next $6$, and so on for $2$ places: $\dfrac{7!}{(7 - 2)!} = 42$.

Common mistakes, and the answer each one gives:

• Used combinations, which ignore the order → $21$
• Arranged all of them rather than only the chosen ones → $5040$

Answer: $15$

A committee is a SET, so order does not matter: $\binom{6}{4} = \dfrac{6!}{4!\,(6 - 4)!} = 15$.

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $360$
• Multiplied the two numbers → $24$