# Conditional Probability and Independence Worksheet ## Section A: Multiple Choice Questions
Q1: A card is drawn from a standard deck of 52 cards. Given that the card is a heart, what is the probability that it is a face card (Jack, Queen, or King)? (a) \(\frac{3}{52}\) (b) \(\frac{3}{13}\) (c) \(\frac{1}{4}\) (d) \(\frac{12}{52}\)
Solution:
Ans: (b) Explanation: This is a conditional probability problem. We need \(P(\text{face card} \mid \text{heart})\). There are 13 hearts in total, and 3 of them are face cards (Jack, Queen, King of hearts). Therefore, the probability is \(\frac{3}{13}\). Option (a) ignores the given condition, option (c) represents the probability of drawing a heart, and option (d) represents the total face cards in the deck without considering the condition.
Q2: Events A and B are independent. If \(P(A) = 0.4\) and \(P(B) = 0.5\), what is \(P(A \cap B)\)? (a) 0.9 (b) 0.2 (c) 0.1 (d) 0.45
Solution:
Ans: (b) Explanation: For independent events, \(P(A \cap B) = P(A) \times P(B)\). Therefore, \(P(A \cap B) = 0.4 \times 0.5 = 0.2\). Option (a) incorrectly adds the probabilities, option (c) is half of the correct answer, and option (d) is the average of the two probabilities.
Q3: In a class of 30 students, 18 play basketball and 12 play soccer. If 8 students play both sports, what is the probability that a randomly selected student plays soccer given that they play basketball? (a) \(\frac{8}{30}\) (b) \(\frac{8}{18}\) (c) \(\frac{8}{12}\) (d) \(\frac{12}{30}\)
Solution:
Ans: (b) Explanation: We need \(P(\text{soccer} \mid \text{basketball})\). Using the conditional probability formula: \(P(S \mid B) = \frac{P(S \cap B)}{P(B)} = \frac{8/30}{18/30} = \frac{8}{18} = \frac{4}{9}\). Option (a) represents \(P(S \cap B)\), option (c) inverts the condition, and option (d) represents \(P(S)\).
Q4: Two dice are rolled. What is the probability of getting a sum of 7 given that the first die shows a 3? (a) \(\frac{1}{36}\) (b) \(\frac{1}{6}\) (c) \(\frac{6}{36}\) (d) \(\frac{1}{3}\)
Solution:
Ans: (b) Explanation: Given the first die shows 3, the second die can show any of 6 values. For the sum to be 7, the second die must show 4. The probability is \(\frac{1}{6}\). Option (a) represents the probability without the condition, option (c) represents the probability of getting a sum of 7 without any condition, and option (d) is double the correct answer.
Q5: Events C and D are mutually exclusive with \(P(C) = 0.3\) and \(P(D) = 0.4\). Are C and D independent? (a) Yes, because they cannot occur together (b) No, because \(P(C \cap D) = 0\) but \(P(C) \times P(D) \neq 0\) (c) Yes, because their probabilities sum to less than 1 (d) Cannot be determined
Solution:
Ans: (b) Explanation:Mutually exclusive events have \(P(C \cap D) = 0\). For independence, we need \(P(C \cap D) = P(C) \times P(D) = 0.3 \times 0.4 = 0.12 \neq 0\). Since these are not equal, the events are not independent. Option (a) confuses mutual exclusivity with independence, option (c) uses an irrelevant criterion, and option (d) is incorrect because we have enough information.
Q6: A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that the second marble is blue given that the first marble was red? (a) \(\frac{3}{8}\) (b) \(\frac{3}{7}\) (c) \(\frac{5}{8}\) (d) \(\frac{2}{7}\)
Solution:
Ans: (b) Explanation: After drawing a red marble first, there are 7 marbles left (4 red and 3 blue). The conditional probability that the second marble is blue is \(\frac{3}{7}\). Option (a) uses the original total, option (c) gives the probability of drawing red second, and option (d) assumes one blue was already drawn.
Q7: Using the multiplication rule for conditional probability, if \(P(A \mid B) = 0.6\) and \(P(B) = 0.25\), what is \(P(A \cap B)\)? (a) 0.85 (b) 0.35 (c) 0.15 (d) 2.4
Solution:
Ans: (c) Explanation: The multiplication rule states \(P(A \cap B) = P(A \mid B) \times P(B)\). Therefore, \(P(A \cap B) = 0.6 \times 0.25 = 0.15\). Option (a) adds the probabilities, option (b) is close but incorrect, and option (d) divides instead of multiplying.
Q8: A fair coin is flipped twice. What is the probability of getting heads on the second flip given that at least one flip was heads? (a) \(\frac{1}{2}\) (b) \(\frac{2}{3}\) (c) \(\frac{1}{3}\) (d) \(\frac{3}{4}\)
Solution:
Ans: (b) Explanation: The sample space given at least one heads is {HH, HT, TH}, which has 3 outcomes. Of these, 2 have heads on the second flip (HH and TH). Therefore, \(P(\text{H on 2nd} \mid \text{at least one H}) = \frac{2}{3}\). Option (a) would be correct without the condition, option (c) gives only one favorable outcome, and option (d) represents a different probability.
## Section B: Fill in the Blanks Q9:The formula for conditional probability \(P(A \mid B)\) is __________, where \(P(B) \neq 0\).
Solution:
Ans: \(\frac{P(A \cap B)}{P(B)}\) Explanation: The conditional probability formula expresses the probability of event A occurring given that event B has already occurred. It is the ratio of the probability of both events occurring to the probability of the given event.
Q10:Two events A and B are said to be __________ if the occurrence of one event does not affect the probability of the occurrence of the other event.
Solution:
Ans: independent Explanation:Independent events satisfy the condition \(P(A \cap B) = P(A) \times P(B)\), or equivalently, \(P(A \mid B) = P(A)\). This means knowing B occurred provides no information about A.
Q11:For two independent events E and F, if \(P(E) = 0.3\) and \(P(F) = 0.7\), then \(P(E \cap F)\) = __________.
Solution:
Ans: 0.21 Explanation: For independent events, we use the multiplication rule: \(P(E \cap F) = P(E) \times P(F) = 0.3 \times 0.7 = 0.21\).
Q12:If events X and Y are mutually exclusive, then \(P(X \cap Y)\) = __________.
Solution:
Ans: 0 Explanation:Mutually exclusive events cannot occur simultaneously. Therefore, the probability of their intersection is zero.
Q13:The probability \(P(A \cup B)\) can be calculated using the formula \(P(A) + P(B) - \)__________.
Solution:
Ans: \(P(A \cap B)\) Explanation: The addition rule for probability states \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). We subtract the intersection to avoid counting it twice.
Q14:If \(P(A \mid B) = P(A)\) and both probabilities are non-zero, then events A and B are __________.
Solution:
Ans: independent Explanation: When the conditional probability of A given B equals the unconditional probability of A, it means B provides no information about A, which is the definition of independence.
## Section C: Word Problems Q15:A medical test for a disease is 95% accurate, meaning it correctly identifies 95% of people who have the disease and correctly identifies 95% of people who do not have the disease. If 2% of the population has the disease, what is the probability that a randomly selected person has the disease given that they tested positive?
Solution:
Ans: Step-by-step solution: Let D = has disease, T = tests positive Given: \(P(D) = 0.02\), \(P(D^c) = 0.98\), \(P(T \mid D) = 0.95\), \(P(T^c \mid D^c) = 0.95\) Therefore: \(P(T \mid D^c) = 0.05\)
Using the law of total probability: \(P(T) = P(T \mid D) \times P(D) + P(T \mid D^c) \times P(D^c)\) \(P(T) = 0.95 \times 0.02 + 0.05 \times 0.98 = 0.019 + 0.049 = 0.068\)
Q16:A box contains 6 defective and 14 non-defective light bulbs. Two bulbs are selected at random without replacement. What is the probability that both bulbs are defective?
Solution:
Ans: Step-by-step solution: Total bulbs = 20 First selection: \(P(\text{defective}) = \frac{6}{20}\) After first defective bulb is selected, 5 defective and 19 total remain Second selection: \(P(\text{defective} \mid \text{first defective}) = \frac{5}{19}\)
Using the multiplication rule: \(P(\text{both defective}) = \frac{6}{20} \times \frac{5}{19} = \frac{30}{380} = \frac{3}{38}\)
Final Answer: \(\frac{3}{38}\) or approximately 0.079
Q17:In a survey of 100 people, 60 like coffee, 50 like tea, and 30 like both. If a person is randomly selected and is known to like coffee, what is the probability that they also like tea?
Solution:
Ans: Step-by-step solution: Let C = likes coffee, T = likes tea Given: \(n(C) = 60\), \(n(T) = 50\), \(n(C \cap T) = 30\) Total people = 100
We need \(P(T \mid C)\) Using the conditional probability formula: \(P(T \mid C) = \frac{P(T \cap C)}{P(C)} = \frac{n(T \cap C)/100}{n(C)/100} = \frac{30}{60} = \frac{1}{2}\)
Final Answer: \(\frac{1}{2}\) or 0.5
Q18:A student takes two independent tests. The probability of passing the first test is 0.8 and the probability of passing the second test is 0.7. What is the probability that the student passes at least one test?
Solution:
Ans: Step-by-step solution: Let A = passes first test, B = passes second test Given: \(P(A) = 0.8\), \(P(B) = 0.7\), A and B are independent
Method 2 - Using addition rule: \(P(A \cap B) = 0.8 \times 0.7 = 0.56\) \(P(A \cup B) = 0.8 + 0.7 - 0.56 = 0.94\)
Final Answer: 0.94
Q19:A jar contains 8 red balls and 4 green balls. Three balls are drawn one after another without replacement. What is the probability that the third ball drawn is red given that the first two balls drawn were red?
Solution:
Ans: Step-by-step solution: Initial: 8 red, 4 green (total 12) After drawing 2 red balls: 6 red, 4 green remain (total 10)
Given the first two were red, we need the conditional probability that the third is red: \(P(\text{3rd red} \mid \text{first 2 red}) = \frac{6}{10} = \frac{3}{5}\)
Final Answer: \(\frac{3}{5}\) or 0.6
Q20:Two dice are rolled simultaneously. Find the probability that the sum is greater than 8 given that both dice show even numbers.
Solution:
Ans: Step-by-step solution: Both dice show even numbers: {2, 4, 6} for each die Possible outcomes when both are even: 3 × 3 = 9 outcomes These are: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)
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