Probability

Answer: $\{2, 3, 4, 5, 10\}$

$A^{c}$ is the elements of $S$ that are NOT in $A\mathclose{:}$ $\{2, 3, 4, 5, 10\}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the complement within $A \cup B$ instead of within $S$ → $\{2, 4\}$

Answer: $\dfrac{20}{51}$

Given that they smoke, only the smokers count: $20$ of the $51$ smokers exercise, so the probability is $\dfrac{20}{51}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by everyone in the survey instead of the smokers → $\dfrac{1}{5}$
• Divided by everyone who exercises instead of the smokers → $\dfrac{4}{7}$

Answer: $\dfrac{671}{1296}$

At least one six is every outcome except no six at all. $P(\text{no six}) = \left(\frac{5}{6}\right)^{4} = \dfrac{625}{1296}\mathclose{,}$ so $P(\text{at least one}) = 1 - \dfrac{625}{1296} = \dfrac{671}{1296}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added $\frac{1}{6}$ once for each roll → $\frac{2}{3}$
• Gave the probability of no sixes → $\frac{625}{1296}$

Answer: $0.001$

Each choice is independent: $0.1^{3} = 0.001\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied $0.1$ by $3$ instead of raising it to the power $3$ → $0.3$
• Used $1 - 0.1$ instead of $0.1$ → $0.729$

Answer: $\{4, 8\}$

$A \setminus B$ is the elements of $A$ that are NOT in $B\mathclose{:}$ $\{4, 8\}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $B \setminus A$ instead → $\{3, 5, 10\}$
• Gave the intersection instead → $\{1, 2\}$

Answer: $0.1$

$A$ is made of the part shared with $B$ and the part outside it: $P(A \setminus B) = P(A) - P(A \cap B) = 0.3 - 0.2 = 0.1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted $P(B)$ instead of $P(A \cap B)$ → $-0.2$
• Gave $P(B \setminus A)$ instead → $0.3$

Answer: $\text{not mutually exclusive}$

$A = \{2, 4, 6\}$ and $B = \{2, 3, 5\}\mathclose{.}$ They share $\{2\}\mathclose{,}$ so they can happen together: not mutually exclusive.

Common mistakes, and the answer each one gives:

• Took events that share an outcome to be mutually exclusive → $\text{mutually exclusive}$

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

There are $6 \times 6 = 36$ equally likely outcomes — $(2, 5)$ and $(5, 2)$ are different ones. Of them, $30$ give a total of at least 5, so the probability is $\dfrac{30}{36} = \dfrac{5}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Counted the 11 possible totals as equally likely → $\frac{8}{11}$
• Counted $(a, b)$ and $(b, a)$ as the same outcome, out of 21 → $\frac{17}{21}$

Answer: $\frac{31}{75}$

Restrict to the $75$ students who take maths; $31$ of them take physics: $\dfrac{31}{75}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by the whole group → $\frac{31}{120}$
• Divided by everyone who takes physics → $\frac{31}{47}$

Answer: $0.3439$

'At least one' is everything except 'none': $1 - 0.9^{4} = 1 - 0.6561 = 0.3439\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took $1 - p^n\mathclose{,}$ the complement of 'all', instead of the complement of 'none' → $0.9999$
• Multiplied $0.1$ by $4$ → $0.4$

Answer: $\{1, 9\}$

$A \cap B$ is the elements in BOTH sets: $\{1, 9\}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the union instead of the intersection → $\{1, 2, 4, 5, 8, 9, 10\}$

Answer: $0.04$

For independent events $P(A \cap B) = P(A) \cdot P(B) = 0.2 \cdot 0.2 = 0.04\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two probabilities → $0.4$
• Gave $P(A \cup B)$ instead → $0.36$

Answer: $0.45$

$P(A \cup B) = P(A) + P(B) - P(A \cap B)\mathclose{:}$ the overlap is in both $A$ and $B\mathclose{,}$ so adding them counts it twice. $0.25 + 0.4 - 0.2 = 0.45\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added $P(A)$ and $P(B)$ without taking away the overlap → $0.65$
• Took the overlap away twice → $0.25$

Answer: $144$

Every outcome of one part can go with every outcome of the others, so multiply: $6 \cdot 6 \cdot 4 = 144\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the numbers of outcomes instead of multiplying → $16$
• Counted the first part only once → $24$

Answer: $\text{independent}$

Independent means $P(A \cap B) = P(A) \cdot P(B)\mathclose{.}$ Here $P(A) \cdot P(B) = 0.6 \cdot 0.8 = 0.48\mathclose{,}$ which IS $P(A \cap B)\mathclose{:}$ independent.

Common mistakes, and the answer each one gives:

• Compared $P(A \cap B)$ with $P(A) + P(B)$ instead of $P(A) \cdot P(B)$ → $\text{not independent}$

Answer: $0.1296$

Each person does not with probability $1 - 0.4 = 0.6\mathclose{,}$ independently: $0.6^{4} = 0.1296\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $0.4$ instead of $1 - 0.4$ → $0.0256$
• Took $1 - p^n$ → $0.9744$

Answer: $64$

Each of the $6$ elements is either in a subset or not: two choices, $6$ times, so $2^{6} = 64$ subsets.

Common mistakes, and the answer each one gives:

• Used $n^2$ instead of $2^n$ → $36$
• Used $2n$ instead of $2^n$ → $12$

Answer: $\dfrac{91}{216}$

At least one six is every outcome except no six at all. $P(\text{no six}) = \left(\frac{5}{6}\right)^{3} = \dfrac{125}{216}\mathclose{,}$ so $P(\text{at least one}) = 1 - \dfrac{125}{216} = \dfrac{91}{216}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added $\frac{1}{6}$ once for each roll → $\frac{1}{2}$
• Gave the probability of no sixes → $\frac{125}{216}$

Answer: $\dfrac{37}{69}$

Given that they smoke, only the smokers count: $37$ of the $69$ smokers exercise, so the probability is $\dfrac{37}{69}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by everyone in the survey instead of the smokers → $\dfrac{37}{103}$
• Divided by everyone who exercises instead of the smokers → $\dfrac{37}{54}$

Answer: $\{1, 2, 3, 6, 7, 9\}$

$A \cup B$ is the elements in EITHER set (or both): $\{1, 2, 3, 6, 7, 9\}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the intersection instead of the union → $\{1, 9\}$

Answer: $0.01$

Each choice is independent: $0.1^{2} = 0.01\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied $0.1$ by $2$ instead of raising it to the power $2$ → $0.2$
• Used $1 - 0.1$ instead of $0.1$ → $0.81$

Answer: $\{2, 3, 8, 9, 10\}$

$A^{c}$ is the elements of $S$ that are NOT in $A\mathclose{:}$ $\{2, 3, 8, 9, 10\}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the complement within $A \cup B$ instead of within $S$ → $\{2\}$

Answer: $\dfrac{1}{6}$

There are $6 \times 6 = 36$ equally likely outcomes — $(2, 5)$ and $(5, 2)$ are different ones. Of them, $6$ give a total of exactly 7, so the probability is $\dfrac{6}{36} = \dfrac{1}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Counted the 11 possible totals as equally likely → $\frac{1}{11}$
• Counted $(a, b)$ and $(b, a)$ as the same outcome, out of 21 → $\frac{1}{7}$

Answer: $\frac{22}{29}$

Restrict to the $145$ students who take maths; $110$ of them take physics: $\dfrac{110}{145} = \frac{22}{29}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by the whole group → $0.55$
• Divided by everyone who takes physics → $\frac{22}{27}$

Answer: $0.4$

$A$ is made of the part shared with $B$ and the part outside it: $P(A \setminus B) = P(A) - P(A \cap B) = 0.5 - 0.1 = 0.4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted $P(B)$ instead of $P(A \cap B)$ → $0.3$
• Gave $P(B \setminus A)$ instead → $0.1$

Answer: $\{1, 5, 7, 9\}$

$A \setminus B$ is the elements of $A$ that are NOT in $B\mathclose{:}$ $\{1, 5, 7, 9\}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $B \setminus A$ instead → $\{2, 10\}$
• Gave the intersection instead → $\{4\}$

Answer: $0.51$

'At least one' is everything except 'none': $1 - 0.7^{2} = 1 - 0.49 = 0.51\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took $1 - p^n\mathclose{,}$ the complement of 'all', instead of the complement of 'none' → $0.91$
• Multiplied $0.3$ by $2$ → $0.6$

Answer: $0.75$

$P(A \cup B) = P(A) + P(B) - P(A \cap B)\mathclose{:}$ the overlap is in both $A$ and $B\mathclose{,}$ so adding them counts it twice. $0.45 + 0.6 - 0.3 = 0.75\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added $P(A)$ and $P(B)$ without taking away the overlap → $1.05$
• Took the overlap away twice → $0.45$

Answer: $0.21$

For independent events $P(A \cap B) = P(A) \cdot P(B) = 0.7 \cdot 0.3 = 0.21\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the two probabilities → $1$
• Gave $P(A \cup B)$ instead → $0.79$

Answer: $\text{not mutually exclusive}$

$A = \{2, 3, 5\}$ and $B = \{3, 6\}\mathclose{.}$ They share $\{3\}\mathclose{,}$ so they can happen together: not mutually exclusive.

Common mistakes, and the answer each one gives:

• Took events that share an outcome to be mutually exclusive → $\text{mutually exclusive}$