Comprehensive Exercises on Random Events and Probability

Approach: Use the classical probability model to compute the probability.

Steps:

1. Sample space: $\Omega = \{(\text{H},\text{H}), (\text{H},\text{T}), (\text{T},\text{H}), (\text{T},\text{T})\}$
2. Event $A = \{\text{exactly one head appears}\} = \{(\text{H},\text{T}), (\text{T},\text{H})\}$
3. $P(A) = \frac{2}{4} = \frac{1}{2}$

Answer: $\frac{1}{2}$

Approach: Compute with the addition formula.

Steps:

1. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
2. $= 0.4 + 0.5 - 0.2 = 0.7$
3. We subtract $P(A \cap B)$ because the intersection was counted twice in the direct addition.

Answer: $P(A \cup B) = 0.7$.

Approach: Compute with the multiplication formula.

Steps:

1. $P(A \cap B) = P(A)P(B|A)$
2. $= 0.3 \times 0.5 = 0.15$
3. Here we multiply by the probability $P(A)=0.3$ that $A$ has already occurred, while $P(B|A)$ is meaningful only under the condition that $A$ occurs.

Answer: $P(A \cap B) = 0.15$, that is, the probability that events $A$ and $B$ occur together is $0.15$.

Approach: Compute with Bayes’ formula.

Steps:

1. Let $A_i = \{\text{box } i \text{ is chosen}\}$, $i = 1, 2, 3$
2. Let $B = \{\text{a white ball is drawn}\}$
3. $P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}$
4. $P(B|A_1) = 1$, $P(B|A_2) = 1$, $P(B|A_3) = \frac{3}{5}$
5. $P(B) = \frac{1}{3} \times 1 + \frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{5} = \frac{11}{15}$
6. $P(A_3|B) = \frac{P(A_3)P(B|A_3)}{P(B)} = \frac{\frac{1}{3} \times \frac{3}{5}}{\frac{11}{15}} = \frac{3}{11}$

Answer: $\frac{3}{11}$

Approach: Check whether the condition for independence is satisfied.

Steps:

1. $P(A)P(B) = 0.6 \times 0.5 = 0.3$
2. $P(A \cap B) = 0.3$
3. Therefore $P(A \cap B) = P(A)P(B)$
4. Hence $A$ and $B$ are independent

Answer: $A$ and $B$ are independent

Approach: Compute with the addition formula.

Steps:

1. Let $A = \{\text{an even number appears}\} = \{2, 4, 6\}$
2. Let $B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}$
3. $A \cap B = \{4, 6\}$
4. $P(A) = \frac{3}{6} = \frac{1}{2}$
5. $P(B) = \frac{3}{6} = \frac{1}{2}$
6. $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$
7. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = \frac{2}{3}$

Answer: $\frac{2}{3}$

Approach: Use the geometric probability model to compute the probability.

Steps:

1. Sample space: $\Omega = [0, 1]$
2. Event $A = [0.3, 0.7]$
3. $m(\Omega) = 1$
4. $m(A) = 0.7 - 0.3 = 0.4$
5. $P(A) = \frac{0.4}{1} = 0.4$

Answer: $0.4$

Approach: Use the classical probability model to compute the probability.

Steps:

1. Sample space: 52 playing cards
2. Event $A = \{\text{a heart is drawn}\}$
3. $|A| = 13$ (there are 13 hearts)
4. $|\Omega| = 52$
5. $P(A) = \frac{13}{52} = \frac{1}{4}$

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

Approach: Compute with Bayes’ formula.

Steps:

1. Let $A_1 = \{\text{the product comes from the first workshop}\}$, $A_2 = \{\text{the product comes from the second workshop}\}$
2. Let $B = \{\text{the product is defective}\}$
3. $P(A_1) = 0.6$, $P(A_2) = 0.4$
4. $P(B|A_1) = 0.01$, $P(B|A_2) = 0.02$
5. $P(B) = 0.6 \times 0.01 + 0.4 \times 0.02 = 0.006 + 0.008 = 0.014$
6. $P(A_1|B) = \frac{0.6 \times 0.01}{0.014} = \frac{0.006}{0.014} = \frac{3}{7}$

Answer: $\frac{3}{7}$

Approach: Compute the relevant probabilities and check independence.

Steps:

1. $P(A) = \frac{3}{6} = \frac{1}{2}$
2. $P(B) = \frac{6}{36} = \frac{1}{6}$ (there are 6 cases: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1))
3. $A \cap B = \{\text{the first die is even and the sum is 7}\} = \{(2,5), (4,3), (6,1)\}$
4. $P(A \cap B) = \frac{3}{36} = \frac{1}{12}$
5. $P(A)P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$
6. Therefore $P(A \cap B) = P(A)P(B)$, and $A$ and $B$ are independent

Answer: $A$ and $B$ are independent

Approach: Use the definition of the complement event and the properties of probability.

Steps:

1. $A \cup \overline{A} = \Omega$
2. $A \cap \overline{A} = \varnothing$
3. Hence $P(A \cup \overline{A}) = P(A) + P(\overline{A})$
4. Also $P(A \cup \overline{A}) = P(\Omega) = 1$
5. Therefore $P(A) + P(\overline{A}) = 1$
6. So $P(\overline{A}) = 1 - P(A)$

Answer: the proof is complete

Approach: Compute from the definitions of the complement event and the difference event.

Steps:

1. $\overline{A} = \Omega - A = \{4, 5\}$
2. $A - B = A \cap \overline{B} = \{1, 2, 3\} \cap \{1, 5\} = \{1\}$

Answer: $\overline{A} = \{4, 5\}$, $A - B = \{1\}$

Approach: Compute from the definitions of events and the rules of event operations.

Steps:

1. $A = \{\text{an even number appears}\} = \{2, 4, 6\}$
2. $B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}$
3. $A \cup B = \{2, 4, 5, 6\}$
4. $A \cap B = \{4, 6\}$

Answer: $A \cup B = \{2, 4, 5, 6\}$, $A \cap B = \{4, 6\}$

Approach: Prove De Morgan’s law through set operations.

Steps:

1. Let $\omega \in \overline{A \cup B}$; then $\omega \notin A \cup B$
2. Therefore $\omega \notin A$ and $\omega \notin B$
3. That is, $\omega \in \overline{A}$ and $\omega \in \overline{B}$
4. Therefore $\omega \in \overline{A} \cap \overline{B}$
5. Conversely, let $\omega \in \overline{A} \cap \overline{B}$; then $\omega \in \overline{A}$ and $\omega \in \overline{B}$
6. Therefore $\omega \notin A$ and $\omega \notin B$
7. That is, $\omega \notin A \cup B$, so $\omega \in \overline{A \cup B}$

Answer: the proof is complete

Approach: Use the distributive law and the properties of event operations.

Steps:

1. $(A \cup B) \cap \overline{A} = (A \cap \overline{A}) \cup (B \cap \overline{A})$
2. $A \cap \overline{A} = \varnothing$
3. Therefore $(A \cup B) \cap \overline{A} = \varnothing \cup (B \cap \overline{A}) = B \cap \overline{A}$

Answer: the proof is complete