Conditional Probability and General Multiplication Rule – Step-by-Step Guidance

Q1. Given that $P(E \text{ and } F) = 0.7$ and $P(E) = 0.8$, what is $P(F|E)$?

Background

Topic: Conditional Probability

This question tests your understanding of how to calculate the probability of event F occurring given that event E has occurred, using the definition of conditional probability.

Key Terms and Formulas

• Conditional Probability: $P(F|E)$ is the probability that event F occurs given that event E has occurred.

• Multiplication Rule: $P(E \text{ and } F) = P(E) \times P(F|E)$

• Conditional Probability Formula: $P(F|E) = \frac{P(E \text{ and } F)}{P(E)}$

Step-by-Step Guidance

1. Identify the known values: $P(E \text{ and } F) = 0.7$, $P(E) = 0.8$.

2. Recall the formula for conditional probability: $P(F|E) = \frac{P(E \text{ and } F)}{P(E)}$.

3. Set up the formula with the given values, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $P(F|E) = 0.875$

We divided $0.7$ by $0.8$ to find the conditional probability.

Q2. Given $P(E) = 0.6$ and $P(F|E) = 0.3$, what is $P(E \text{ and } F)$?

Background

Topic: General Multiplication Rule

This question tests your ability to use the multiplication rule for probabilities, especially when conditional probability is given.

Key Terms and Formulas

• Multiplication Rule: $P(E \text{ and } F) = P(E) \times P(F|E)$

• Conditional Probability: $P(F|E)$ is the probability of F given E.

Step-by-Step Guidance

1. Identify the known values: $P(E) = 0.6$, $P(F|E) = 0.3$.

2. Recall the multiplication rule: $P(E \text{ and } F) = P(E) \times P(F|E)$.

3. Set up the multiplication using the given values, but do not multiply yet.

Try solving on your own before revealing the answer!

Final Answer: $P(E \text{ and } F) = 0.18$

We multiplied $0.6$ by $0.3$ to get the joint probability.

Q3. In 25% of marriages, the woman has a bachelor's degree and the marriage lasts at least 20 years. 26% of women have a bachelor's degree. What is the probability a randomly selected marriage will last at least 20 years if the woman has a bachelor's degree?

Background

Topic: Conditional Probability in Context

This question asks you to interpret real-world data using conditional probability, specifically the probability that a marriage lasts at least 20 years given the woman has a bachelor's degree.

Key Terms and Formulas

• Let A = "woman has a bachelor's degree" and B = "marriage lasts at least 20 years".

• Conditional Probability: $P(B|A) = \frac{P(A \text{ and } B)}{P(A)}$

Step-by-Step Guidance

1. Assign events: Let A = woman has a bachelor's degree, B = marriage lasts at least 20 years.

2. Identify the given probabilities: $P(A \text{ and } B) = 0.25$, $P(A) = 0.26$.

3. Set up the conditional probability formula: $P(B|A) = \frac{P(A \text{ and } B)}{P(A)}$.

4. Plug in the values, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $P(B|A) \approx 0.96$

Dividing $0.25$ by $0.26$ gives the probability that a marriage lasts at least 20 years given the woman has a bachelor's degree.

Q4(a). A bag contains 29 tulip bulbs: 10 red, 9 yellow, 10 purple. Two bulbs are selected without replacement. What is the probability both are red?

Background

Topic: Probability Without Replacement

This question tests your understanding of dependent probability, where the outcome of the first selection affects the probability of the second.

Key Terms and Formulas

• Probability without replacement: The total number of bulbs decreases after the first selection.

• Multiplication Rule for Dependent Events: $P(A \text{ and } B) = P(A) \times P(B|A)$

Step-by-Step Guidance

1. Find the probability the first bulb is red: $P_1 = \frac{10}{29}$.

2. After removing one red bulb, there are 9 red bulbs left and 28 bulbs total. Probability the second bulb is red: $P_2 = \frac{9}{28}$.

3. Multiply the probabilities: $P(\text{both red}) = P_1 \times P_2$.

Try solving on your own before revealing the answer!

Final Answer: $P(\text{both red}) = \frac{10}{29} \times \frac{9}{28} \approx 0.110$

We multiplied the probability of picking a red bulb first by the probability of picking another red bulb second, without replacement.

Q4(b). What is the probability that the first bulb selected is red and the second is yellow?

Background

Topic: Probability Without Replacement (Different Colors)

This question tests your ability to calculate the probability of two dependent events with different outcomes.

Key Terms and Formulas

• Probability without replacement: The total number of bulbs decreases after the first selection.

• Multiplication Rule for Dependent Events: $P(A \text{ and } B) = P(A) \times P(B|A)$

Step-by-Step Guidance

1. Probability the first bulb is red: $P_1 = \frac{10}{29}$.

2. After removing one red bulb, there are still 9 yellow bulbs and 28 bulbs total. Probability the second bulb is yellow: $P_2 = \frac{9}{28}$.

3. Multiply the probabilities: $P(\text{red then yellow}) = P_1 \times P_2$.

Try solving on your own before revealing the answer!

Final Answer: $P(\text{red then yellow}) = \frac{10}{29} \times \frac{9}{28} \approx 0.110$

We used the multiplication rule for dependent events, considering the change in total bulbs after the first draw.

Q5. A professor randomly selects two people from a class roster (Jinita, Dave, Mike, Al, William, Pam, Allison, Neta, Jim, Kristin). What is the probability that Al is chosen first and Jinita is chosen second?

Background

Topic: Probability of Ordered Selection Without Replacement

This question tests your understanding of probability when selecting individuals in a specific order without replacement.

Key Terms and Formulas

• Probability of first event: $P_1 = \frac{1}{10}$ (since there are 10 people).

• Probability of second event (after first is removed): $P_2 = \frac{1}{9}$.

• Multiplication Rule: $P(\text{Al first and Jinita second}) = P_1 \times P_2$.

Step-by-Step Guidance

1. Probability Al is chosen first: $P_1 = \frac{1}{10}$.

2. After Al is chosen, 9 people remain. Probability Jinita is chosen second: $P_2 = \frac{1}{9}$.

3. Multiply the probabilities: $P = P_1 \times P_2$.

Try solving on your own before revealing the answer!

Final Answer: $P = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \approx 0.011$

We multiplied the probability of selecting Al first by the probability of selecting Jinita second from the remaining students.

Q6(a). Among fatal crashes in normal weather, what is the probability that a randomly selected fatal crash occurs when it is daylight?

Background

Topic: Conditional Probability from a Two-Way Table

This question tests your ability to use a contingency table to find conditional probabilities.

Key Terms and Formulas

• Conditional Probability: $P(\text{Daylight} | \text{Normal Weather}) = \frac{\text{Number in Daylight and Normal Weather}}{\text{Total in Normal Weather}}$

Step-by-Step Guidance

1. Find the total number of fatal crashes in normal weather by summing all light conditions for normal weather.

2. Identify the number of fatal crashes in daylight and normal weather (given as 14,307).

3. Set up the conditional probability formula: $P = \frac{14,307}{\text{Total in Normal Weather}}$.

Try solving on your own before revealing the answer!

Final Answer: $P \approx 0.429$

We divided the number of daylight/normal weather crashes by the total number of normal weather crashes.

Q6(b). Among fatal crashes when it is daylight, what is the probability that a randomly selected fatal crash occurs in normal weather?

Background

Topic: Conditional Probability from a Two-Way Table (Reversed Condition)

This question tests your ability to reverse the condition in a conditional probability problem using a contingency table.

Key Terms and Formulas

• Conditional Probability: $P(\text{Normal Weather} | \text{Daylight}) = \frac{\text{Number in Daylight and Normal Weather}}{\text{Total in Daylight}}$

Step-by-Step Guidance

1. Find the total number of fatal crashes in daylight by summing all weather conditions for daylight.

2. Identify the number of fatal crashes in daylight and normal weather (14,307).

3. Set up the conditional probability formula: $P = \frac{14,307}{\text{Total in Daylight}}$.

Try solving on your own before revealing the answer!

Final Answer: $P \approx 0.902$

We divided the number of daylight/normal weather crashes by the total number of daylight crashes.

Q6(c). Is 'dark, but lighted' more dangerous in normal weather or in snow/sleet? Explain.

Background

Topic: Comparing Relative Risk

This question asks you to compare the relative risk (probability) of fatal crashes in 'dark, but lighted' conditions for two different weather types.

Key Terms and Formulas

• Relative Risk: Compare $P(\text{Dark, but Lighted} | \text{Normal Weather})$ and $P(\text{Dark, but Lighted} | \text{Snow/Sleet})$.

• Conditional Probability: $P = \frac{\text{Number in Dark, but Lighted and Weather}}{\text{Total in Weather}}$

Step-by-Step Guidance

1. Find the number of fatal crashes in 'dark, but lighted' for normal weather (8,151) and for snow/sleet (156).

2. Find the total number of fatal crashes for normal weather and for snow/sleet by summing across all light conditions for each weather type.

3. Set up the conditional probability for each: $P_{\text{normal}} = \frac{8,151}{\text{Total Normal}}$, $P_{\text{snow}} = \frac{156}{\text{Total Snow/Sleet}}$.

4. Compare the two probabilities to determine which is higher.

Try solving on your own before revealing the answer!

Final Answer: 'Dark, but lighted' is more dangerous in snow/sleet (higher proportion of crashes occur in this condition).

The probability is higher for snow/sleet when you compare the proportions.