Combinatorics, random variables and distributions

Answer: $75$

Every choice of the first can go with every choice of the second, and each pair with every choice of the third: $5 \cdot 5 \cdot 3 = 75\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the numbers of choices instead of multiplying → $13$
• Multiplied by $3!$ as well, as if the order of the three mattered → $450$

Answer: $16\%$

$185$ is $1$ standard deviation above the mean. $68\%$ lie within $1$ of the mean, so $32\%$ lie outside, split equally between the two tails: $16\%\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the percentage inside the band instead of beyond it → $68\%$
• Did not halve: counted both tails → $32\%$

Answer: $0.8$

Add the probabilities of every value up to and including $1\mathclose{:}$ $0.2 + 0.1 + 0.5 = 0.8\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out $x = 1$ itself → $0.3$
• Gave only $P(X = 1)$ → $0.5$

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

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

Common mistakes, and the answer each one gives:

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

Answer: $10$

A committee is a SET, so order does not matter: $\binom{5}{3} = \dfrac{5!}{3!\,(5 - 3)!} = 10\mathclose{.}$

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $60$
• Divided $5!$ by $3!$ only, forgetting $(5 - 3)!$ → $20$

Answer: $980$

$x = \mu + z\sigma = 1000 + (-1) \cdot 20 = 980$ g.

Common mistakes, and the answer each one gives:

• Subtracted $z\sigma$ instead of adding it → $1020$
• Added $z$ itself, without multiplying by $\sigma$ → $999$

Answer: $30$

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

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: $16\%$

$51$ is $1$ standard deviation below the mean. $68\%$ lie within $1$ of the mean, so $32\%$ lie outside, split equally between the two tails: $16\%\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the percentage inside the band instead of beyond it → $68\%$
• Did not halve: counted both tails → $32\%$

Answer: $2.3$

Multiply each value by its probability and add: $E(X) = (-3) \cdot 0.1 + (-2) \cdot 0.5 + 9 \cdot 0.4 = 2.3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Averaged the values, ignoring the probabilities → $\frac{4}{3}$

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

Binomial with $n = 2$ and $p = \frac{1}{6}\mathclose{:}$ $\binom{2}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right) = 2 \times \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{18}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the factor for the rolls that are NOT sixes → $\frac{1}{3}$
• Left out $\binom{2}{1}\mathclose{,}$ the number of orders → $\frac{5}{36}$

Answer: $90$

Choose the women and the men separately, and multiply — every choice of women goes with every choice of men: $\binom{6}{2} \times \binom{6}{1} = 15 \times 6 = 90\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two counts instead of multiplying them → $21$
• Chose $3$ people from all $12$ at once → $220$

Answer: $-0.5$

$z = \dfrac{x - \mu}{\sigma} = \dfrac{987.5 - 1000}{25} = -0.5\mathclose{:}$ $0.5$ standard deviations below the mean.

Common mistakes, and the answer each one gives:

• Divided by the variance instead of the standard deviation → $-0.02$
• Subtracted the wrong way round → $0.5$

Answer: $180$

LETTER has $6$ letters, so $6!$ orders if they were all different. Swapping two equal letters changes nothing, so divide by $2! \cdot 2!$ for the repeats: $\dfrac{6!}{2! \cdot 2!} = 180\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by the repeated letters → $720$
• Divided by the repeats of only one letter → $360$

Answer: $34\%$

From the mean up to $1$ standard deviation above it is half of the $68\%$ band, because the curve is symmetric: $34\%\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not halve: used the whole band either side of the mean → $68\%$

Answer: $6$

The probabilities add to 1, so $p = 1 - 0.2 - 0.4 = 0.4\mathclose{.}$ Then $E(X) = 1 \cdot 0.4 + 8 \cdot 0.2 + 10 \cdot 0.4 = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Averaged the values, ignoring the probabilities → $\frac{19}{3}$
• Left the unknown probability out, as if it were 0 → $5.6$

Answer: $0.441$

Binomial with $n = 3$ and $p = 0.3\mathclose{:}$ $\binom{3}{1} \cdot 0.3^{1} \cdot 0.7^{2} = 3 \cdot 0.3 \cdot 0.49 = 0.441\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out $\binom{n}{k}\mathclose{,}$ the number of orders → $0.147$
• Swapped $p$ and $1 - p$ → $0.189$

Answer: $126$

Every choice of the first can go with every choice of the second, and each pair with every choice of the third: $7 \cdot 6 \cdot 3 = 126\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the numbers of choices instead of multiplying → $16$
• Multiplied by $3!$ as well, as if the order of the three mattered → $756$

Answer: $99.7\%$

$350$ and $650$ are $3$ standard deviations either side of the mean, and the rule says $99.7\%$ lie within that band.

Common mistakes, and the answer each one gives:

• Used the percentage for a different number of standard deviations → $95\%$
• Halved it, as if only one side counted → $49.85\%$

Answer: $15$

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

Common mistakes, and the answer each one gives:

• Counted the orders as different, which a committee does not → $360$
• Divided $6!$ by $4!$ only, forgetting $(6 - 4)!$ → $30$

Answer: $52$

$x = \mu + z\sigma = 60 + (-1) \cdot 8 = 52$ points.

Common mistakes, and the answer each one gives:

• Subtracted $z\sigma$ instead of adding it → $68$
• Added $z$ itself, without multiplying by $\sigma$ → $59$

Answer: $0.9$

Add the probabilities of every value up to and including $6\mathclose{:}$ $0.4 + 0.5 = 0.9\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out $x = 6$ itself → $0.4$
• Gave only $P(X = 6)$ → $0.5$

Answer: $\dfrac{1}{4}$

Each of the $2^{4} = 16$ sequences is equally likely, and $\binom{4}{3} = 4$ of them have exactly $3$ heads. So the probability is $\dfrac{4}{16} = \dfrac{1}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

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

Answer: $840$

The first place has $7$ candidates, the next $6\mathclose{,}$ and so on for $4$ places: $7 \times 6 \times 5 \times 4 = 840\mathclose{,}$ which is $\dfrac{7!}{(7 - 4)!}\mathclose{.}$

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: $0.15\%$

$1030$ is $3$ standard deviations above the mean. $99.7\%$ lie within $3$ of the mean, so $0.3\%$ lie outside, split equally between the two tails: $0.15\%\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the percentage inside the band instead of beyond it → $99.7\%$
• Did not halve: counted both tails → $0.3\%$

Answer: $60$

Choose the women and the men separately, and multiply — every choice of women goes with every choice of men: $\binom{4}{2} \times \binom{5}{3} = 6 \times 10 = 60\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two counts instead of multiplying them → $16$
• Chose $5$ people from all $9$ at once → $126$

Answer: $1.5$

$z = \dfrac{x - \mu}{\sigma} = \dfrac{185 - 170}{10} = 1.5\mathclose{:}$ $1.5$ standard deviations above the mean.

Common mistakes, and the answer each one gives:

• Divided by the variance instead of the standard deviation → $0.15$
• Subtracted the wrong way round → $-1.5$

Answer: $6.1$

Multiply each value by its probability and add: $E(X) = 5 \cdot 0.5 + 6 \cdot 0.2 + 8 \cdot 0.3 = 6.1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Averaged the values, ignoring the probabilities → $\frac{19}{3}$

Answer: $\dfrac{5}{324}$

Binomial with $n = 4$ and $p = \frac{1}{6}\mathclose{:}$ $\binom{4}{3}\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right) = 4 \times \dfrac{1}{216} \times \dfrac{5}{6} = \dfrac{5}{324}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the factor for the rolls that are NOT sixes → $\frac{1}{54}$
• Left out $\binom{4}{3}\mathclose{,}$ the number of orders → $\frac{5}{1296}$

Answer: $20$

ADDED has $5$ letters, so $5!$ orders if they were all different. Swapping two equal letters changes nothing, so divide by $3!$ for the repeats: $\dfrac{5!}{3!} = 20\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not divide by the repeated letters → $120$

Answer: $2.5\%$

$1000$ is $2$ standard deviations below the mean. $95\%$ lie within $2$ of the mean, so $5\%$ lie outside, split equally between the two tails: $2.5\%\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the percentage inside the band instead of beyond it → $95\%$
• Did not halve: counted both tails → $5\%$