Hard GRE Conditional Probability Questions Practice Questions

Conditional probability measures the likelihood of an event occurring given that another event has already taken place. This mathematical concept is a pillar of the GRE Prep curriculum, often appearing in the Quantitative Reasoning section to test a student's ability to filter sample spaces based on specific constraints.

Concept Explanation

Conditional probability is the probability of an event $A$ occurring, given that event $B$ has already occurred, and it is calculated by dividing the probability of both events happening by the probability of the given condition. In formal notation, this is expressed as: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ Where $P(A|B)$ represents the "probability of $A$ given $B$," $P(A \cap B)$ is the probability that both events occur simultaneously, and $P(B)$ is the probability of the condition. On the GRE, these problems often involve dependent events where the outcome of the first trial changes the total number of possibilities for the second trial. A common way to visualize these problems is through a contingency table or a tree diagram. For instance, if you are drawing marbles from a bag without replacement, the total number of marbles decreases after the first draw. This shift in the denominator is the essence of conditional logic. According to Khan Academy, understanding the restriction of the sample space is the most effective way to solve these problems without relying solely on formulas. When you are presented with Hard GRE Conditional Probability Questions, the difficulty usually stems from multi-step dependencies or the inclusion of combinations and permutations within the probability calculation.

Solved Examples

1. Example 1: The Two-Child Problem
   A family has two children. Given that at least one of the children is a girl, what is the probability that both children are girls? Assume the probability of having a boy or girl is equal.
   1. Identify the initial sample space for two children: {BB, BG, GB, GG}. Each has a probability of $\frac{1}{4}$.
   2. Apply the condition: "At least one is a girl." This eliminates the BB option. The new sample space is {BG, GB, GG}.
   3. Find the target event: Both are girls (GG).
   4. Calculate the probability: There is 1 favorable outcome out of 3 possible outcomes in the restricted space. The answer is $\frac{1}{3}$.
2. Example 2: Dependent Drawing
   A box contains 5 red balls and 3 blue balls. If two balls are drawn one after another without replacement, what is the probability that the second ball is blue given that the first ball was red?
   1. Identify the condition: The first ball drawn is red.
   2. Update the sample space: Initially, there were 8 balls. After removing one red ball, 7 balls remain (4 red, 3 blue).
   3. Identify the target: Drawing a blue ball from the remaining set.
   4. Calculate: The probability is the number of blue balls divided by the new total: $\frac{3}{7}$.
3. Example 3: Disease Testing
   In a population, 1% of people have a specific disease. A test for the disease is 99% accurate (it returns a positive result 99% of the time for those with the disease and a negative result 99% of the time for those without it). If a person tests positive, what is the probability they actually have the disease?
   1. Use a hypothetical population of 10,000. 100 people have the disease (1%), and 9,900 do not.
   2. Calculate true positives: 99% of 100 = 99 people.
   3. Calculate false positives: 1% of 9,900 = 99 people.
   4. Total positive tests = $99 + 99 = 198$.
   5. Probability of disease given positive test: $\frac{99}{198} = \frac{1}{2}$ or 50%.

Practice Questions

1. A jar contains 4 black marbles and 6 white marbles. Two marbles are chosen at random without replacement. What is the probability that the second marble is black, given that the first marble was white?
2. In a class of 30 students, 20 like math, 15 like science, and 10 like both. If a student is chosen at random and they like math, what is the probability they also like science?
3. A fair six-sided die is rolled. Given that the result is an even number, what is the probability that the number is a multiple of 3?

4. Two cards are drawn from a standard 52-card deck without replacement. What is the probability that both are Aces, given that at least one is an Ace?
5. A manufacturing plant has two machines, A and B. Machine A produces 60% of the items and Machine B produces 40%. 5% of Machine A’s items are defective, and 10% of Machine B’s items are defective. If an item is randomly selected and found to be defective, what is the probability it came from Machine B?
6. A bag contains 3 red, 4 green, and 5 blue balls. Three balls are drawn without replacement. What is the probability that the third ball is red, given that the first two were green?
7. In a survey of 100 people, 60 drink coffee, 40 drink tea, and 20 drink both. If a person is selected who does not drink tea, what is the probability they drink coffee?
8. Three fair coins are tossed. Given that at least two coins show heads, what is the probability that all three coins show heads?
9. A drawer contains 10 socks: 6 are blue and 4 are red. If two socks are drawn at random, what is the probability the second is red given the first was red?
10. A city weather report states there is a 40% chance of rain. If it rains, there is an 80% chance of traffic jams. If it does not rain, there is only a 30% chance of traffic jams. Given that there is a traffic jam, what is the probability it rained?

Answers & Explanations

1. Answer: $\frac{4}{9}$. After drawing one white marble, the total number of marbles drops to 9. The number of black marbles remains 4. Therefore, $P = \frac{4}{9}$.
2. Answer: $\frac{1}{2}$. The condition is "likes math" (20 students). Out of these 20, 10 also like science. $P = \frac{10}{20} = \frac{1}{2}$.
3. Answer: $\frac{1}{3}$. The even numbers on a die are {2, 4, 6}. Out of these 3 possibilities, only 6 is a multiple of 3. Thus, $P = \frac{1}{3}$.
4. Answer: $\frac{1}{33}$. Total pairs with at least one Ace = $\binom{52}{2} - \binom{48}{2} = 1326 - 1128 = 198$. Pairs with two Aces = $\binom{4}{2} = 6$. Probability = $\frac{6}{198} = \frac{1}{33}$.
5. Answer: $\frac{4}{7}$. Defective from A: $0.60 \times 0.05 = 0.03$. Defective from B: $0.40 \times 0.10 = 0.04$. Total defective = 0.07. $P(B|Defective) = \frac{0.04}{0.07} = \frac{4}{7}$.
6. Answer: $\frac{3}{10}$. After 2 green balls are removed, the total is $12 - 2 = 10$. The number of red balls is still 3. $P = \frac{3}{10}$.
7. Answer: $\frac{2}{3}$. Total people = 100. Those who drink tea = 40. Those who do not drink tea = 60. Of the 60 coffee drinkers, 20 drink tea, so 40 drink only coffee. $P = \frac{40}{60} = \frac{2}{3}$.
8. Answer: $\frac{1}{4}$. Sample space for at least 2 heads: {HHH, HHT, HTH, THH}. There are 4 outcomes. Only 1 (HHH) has three heads. $P = \frac{1}{4}$.
9. Answer: $\frac{3}{9} = \frac{1}{3}$. After one red sock is removed, 9 socks remain, 3 of which are red. $P = \frac{3}{9}$.
10. Answer: $\frac{32}{50} = 0.64$. $P( \text{Rain} \cap \text{Traffic}) = 0.4 \times 0.8 = 0.32$. $P( \text{No Rain} \cap \text{Traffic}) = 0.6 \times 0.3 = 0.18$. Total Traffic = 0.50. $P = \frac{0.32}{0.50} = 0.64$.

1. If \( P(A) = 0.5 \), \( P(B) = 0.4 \), and \( P(A \cap B) = 0.2 \), what is \( P(A|B) \)?

• A0.25
• B0.4
• C0.5
• D0.8

Frequently Asked Questions

What is the difference between independent and dependent probability?

Independent events do not influence each other's outcomes, whereas dependent events occur when the first event changes the probability of the second. Conditional probability is most useful for calculating outcomes in dependent scenarios where the sample space is restricted.

How do I identify a conditional probability question on the GRE?

Look for keywords like "given that," "if," or "knowing that," which signal a restriction on the total outcomes. For more practice identifying these, check out GRE practice questions with explanations.

Can conditional probability be greater than 1?

No, like all probabilities, conditional probability must range between 0 and 1, inclusive. If your calculation results in a number greater than 1, you likely inverted the numerator and denominator in the formula.

Is the order of events important in conditional probability?

Yes, $P(A|B)$ is generally not equal to $P(B|A)$ unless the individual probabilities of A and B are the same. This distinction is vital for accurately solving complex word problems in the GRE Quantitative section.

What is the best way to study for hard probability questions?

Consistent practice with varied problem sets is essential for mastering these concepts. You can use an AI Exam Simulator to generate realistic test scenarios that focus specifically on your weak areas in probability.

Do I need to memorize Bayes' Theorem for the GRE?

While the full theorem is rarely strictly required, understanding the logic of "Total Probability" as a denominator is very helpful for high-difficulty questions. Most GRE problems can be solved by carefully adjusting the sample space manually.