Applying the Multiplication Rule for Probability - Statistics and Probability

A medical screening is done in two stages. First, a person is randomly selected from a group; the probability the person is in the high-risk category is $P(A)=0.15$ . Then, given the person is high-risk, the probability the test result is positive is $P(B\mid A)=0.80$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $0.15\times 0.80=0.12$ This question tests the Multiplication Rule in medical screening. The "both" indicates we need P(A and B), meaning being high-risk AND testing positive. First, a person is high-risk with probability P(A) = 0.15. Then, given they are high-risk, the conditional probability of testing positive is P(B|A) = 0.80. By the Multiplication Rule, P(A ∩ B) = P(A) × P(B|A) = 0.15 × 0.80 = 0.12. A common mistake is adding the probabilities (0.15 + 0.80 = 0.95), but we multiply because testing positive is conditional on being high-risk.
A card is drawn first from a standard 52-card deck and kept out (no replacement). Event $A$ is "the first card is a heart," and event $B$ is "the second card is a heart." Given $P(A)=\frac{13}{52}$ and $P(B\mid A)=\frac{12}{51}$ , what is the probability that both events occur, $P(A\cap B)$ ?
Answer: $\frac{13}{52}\cdot\frac{12}{51}=\frac{1}{17}$ This question tests the Multiplication Rule for Probability, used to find the probability of both events in a dependent sequence without replacement. It treats 'and' as sequential: first draw event A, then event B conditional on A. Here, event A is drawing a heart first from a 52-card deck, with $P(A) = \frac{13}{52}$ . Given the first was a heart, the conditional probability of drawing another heart (event B) is $P(B\mid A) = \frac{12}{51}$ , as one heart is removed. Multiplying yields $P(A \cap B) = \frac{13}{52} \times \frac{12}{51} = \frac{1}{17}$ , the probability of drawing two hearts in sequence without replacement. A common error is assuming independence and using $\frac{13}{52} \times \frac{13}{52}$ , but this overlooks the deck change after the first draw. A helpful strategy is to read it as 'first a heart, then another heart given the first was removed.'
A website user session proceeds in two steps. A user first clicks an ad (event $A$ ). If the user clicks the ad, the user then makes a purchase (event $B$ ). If $P(A)=0.20$ and $P(B\mid A)=0.35$ , what is the probability that both $A$ and $B$ occur (i.e., $P(A\cap B)$ )?
Answer: $0.07$ This question tests the multiplication rule for probability, essential for finding the probability of sequential events both occurring. The 'and' signifies a conditional sequence: first the user clicks the ad (A), then makes a purchase (B) given they clicked. Event A is clicking the ad, with P(A) = 0.20. The conditional P(B|A) = 0.35 is the probability of purchasing given a click. Multiplying yields P(A ∩ B) = 0.20 × 0.35 = 0.07, the overall probability of both clicking and purchasing in a session. A frequent error is treating events as independent and ignoring the conditioning, but here B is explicitly conditional on A. A helpful strategy is to interpret 'A and B' as 'first A, then B given A,' which aligns with this website user behavior context.
A manufacturing line has a two-step outcome. First, an item is assembled by Robot 1 (event $A$ ) with probability $P(A)=0.55$ . Then, given it was assembled by Robot 1, it is packaged correctly (event $B$ ) with probability $P(B\mid A)=0.92$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $0.506$ This question tests the Multiplication Rule for Probability, computing the joint probability in a manufacturing line with dependent steps. It treats 'and' as a sequence: first event A, then event B conditional on A. Event A is assembly by Robot 1, with P(A) = 0.55. Given Robot 1 assembly, the conditional probability of correct packaging (event B) is P(B|A) = 0.92. Multiplying gives P(A ∩ B) = 0.55 × 0.92 = 0.506, the probability of both Robot 1 assembly and correct packaging. A usual error is assuming independence and squaring or adding, but packaging depends on the assembler. Parse as 'first assembled by Robot 1, then packaged correctly given that.'
A factory produces light bulbs. A bulb is selected at random and first tested for being defective (event $A$ ). If it is defective, it is then checked to see whether the defect is visible (event $B$ ). If $P(A)=0.06$ and $P(B\mid A)=0.70$ , what is the probability that both $A$ and $B$ occur (i.e., $P(A\cap B)$ )?
Answer: $0.042$ This question tests the multiplication rule for probability, which is used to find the probability of two events both occurring. The rule states that the probability of A and B happening is P(A) multiplied by the conditional probability P(B given A), reflecting the sequential nature where 'and' means first A occurs, then B given that A has happened. Here, the first event A is selecting a defective bulb, with P(A) = 0.06. The second event B is the defect being visible given that it's defective, with P(B|A) = 0.70. Multiplying these gives P(A ∩ B) = 0.06 × 0.70 = 0.042, which is the probability that a randomly selected bulb is both defective and has a visible defect. A common mistake is assuming independence and just multiplying P(A) and P(B) without conditioning, but since B depends on A, we must use the conditional probability. A good strategy is to read 'A and B' as 'first A happens, then B given A,' which helps apply the multiplication rule correctly in this factory quality control context.
A factory inspects a batch of parts in two steps. First, a part is checked for a surface defect (event $A$ ). If it passes that check, it is then tested for electrical performance (event $B$ ). Suppose $P(A)=0.80$ and $P(B\mid A)=0.95$ . What is the probability that both $A$ and $B$ occur (i.e., the part passes the surface check and then passes the electrical test), $P(A\cap B)$ ?
Answer: $0.76$ This question tests the Multiplication Rule for probability. The word "and" here means we need the part to pass the surface check first, then pass the electrical test - these events happen sequentially. The first event A (passing surface check) has probability P(A) = 0.80. Given that the part passed the surface check, the conditional probability of passing the electrical test is P(B|A) = 0.95. By the Multiplication Rule, P(A∩B) = P(A) × P(B|A) = 0.80 × 0.95 = 0.76. A common mistake would be to add the probabilities (getting 1.75) or assume the events are independent without using the given conditional probability.
A factory line has a two-stage check. First, an item is selected; $P(A)=0.18$ that it comes from Machine 1. Then, given it comes from Machine 1, the probability it fails the final check is $P(B\mid A)=0.10$ . What is the probability that both the item comes from Machine 1 and it fails the final check (i.e., $P(A\cap B)$ )?
Answer: $0.18\times 0.10=0.018$ This question tests the Multiplication Rule for Probability, crucial for dependent events in checks. The 'and' indicates both coming from Machine 1 and failing the check, sequentially with failure conditional on the machine. Event A is item from Machine 1, with P(A) = 0.18. Event B is failing the check given from Machine 1, with P(B|A) = 0.10. So, P(A ∩ B) = 0.18 × 0.10 = 0.018, the probability of both in the factory line. A typical mistake is adding or using unrelated probabilities, assuming independence, but failure rates depend on the machine. To avoid errors, interpret 'and' as 'first from Machine 1, then fails check,' emphasizing conditionality in production.
A board game has two sequential outcomes. First, you roll a 6 (event $A$ ). Given that you rolled a 6, you then draw a "bonus" card from a deck where 3 out of 10 cards are bonus cards (event $B$ ). If $P(A)=\frac{1}{6}$ and $P(B\mid A)=\frac{3}{10}$ , what is $P(A\cap B)$ , the probability that both $A$ and $B$ occur?
Answer: $\frac{1}{20}$ This question tests the Multiplication Rule for probability. The "and" indicates sequential events in the game: first roll a 6, then draw a bonus card. The first event A (rolling a 6) has probability P(A) = 1/6. Given that you rolled a 6, the conditional probability of drawing a bonus card is P(B|A) = 3/10. By the Multiplication Rule, P(A∩B) = P(A) × P(B|A) = (1/6) × (3/10) = 3/60 = 1/20. A common mistake is assuming these events affect each other when they don't - the die roll doesn't change the card probabilities, but we still multiply because we need both to happen.
A warehouse does a two-step quality check. First, an item is chosen from Supplier X with probability $P(A)=0.25$ . Then, if it came from Supplier X, it is defective with probability $P(B\mid A)=0.08$ . What is the probability that both events occur (chosen from Supplier X and defective), $P(A\cap B)$ ?
Answer: $0.25\times 0.08=0.02$ This question tests the Multiplication Rule for quality control events. The "and" means we need both choosing from Supplier X AND the item being defective. First, an item is chosen from Supplier X with probability P(A) = 0.25. Then, given it came from Supplier X, the conditional probability it's defective is P(B|A) = 0.08. By the Multiplication Rule, P(A ∩ B) = P(A) × P(B|A) = 0.25 × 0.08 = 0.02. A common error is assuming independence and using 0.25 × 0.25, but the defect rate specifically depends on coming from Supplier X.
A card is drawn from a standard 52-card deck first and not replaced. Event $A$ is "the first card is a heart," so $P(A)=\frac{13}{52}$ . Event $B$ is "the second card is a king." Given that the first card was a heart, $P(B\mid A)=\frac{4}{51}$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $\frac{13}{52}\cdot\frac{4}{51}=\frac{1}{51}$ This question tests the Multiplication Rule for probability. The process involves drawing two cards sequentially without replacement: first a heart, then a king. The first event is drawing a heart, with probability P(A) = 13/52. The second event is conditional: given that the first card was a heart (and not replaced), drawing a king has probability P(B|A) = 4/51. To find P(A∩B), we multiply: (13/52) × (4/51) = 52/2652 = 1/51. A common mistake is to use 4/52 for the second probability, forgetting that we now have only 51 cards. The strategy is to track how the first draw affects the second.
A student applies for a scholarship in two steps. First, their application is submitted on time (event $A$ ) with probability $P(A)=0.85$ . Then, given it was submitted on time, it is accepted for review (event $B$ ) with probability $P(B\mid A)=0.40$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $0.34$ This question tests the Multiplication Rule for Probability, to find the joint probability in a two-step scholarship application process. The rule interprets 'and' as sequential and conditional: first event A, then event B given A. Event A is submitting on time, with P(A) = 0.85. Given on-time submission, the conditional probability of acceptance for review (event B) is P(B|A) = 0.40. Thus, P(A ∩ B) = 0.85 × 0.40 = 0.34, the probability of both timely submission and acceptance. People frequently mistake this for independent events and add probabilities, but review depends on timeliness. A strategy is to read it as 'first on time, then accepted given on time.'
A card is drawn from a standard 52-card deck without replacement, and then a second card is drawn. Let event $A$ be "the first card is a heart," so $P(A)=\frac{13}{52}$ . Let event $B$ be "the second card is a heart given the first card was a heart," so $P(B\mid A)=\frac{12}{51}$ . What is the probability that both cards drawn are hearts (i.e., $P(A\cap B)$ )?
Answer: $\frac{13}{52}\cdot\frac{12}{51}=\frac{1}{17}$ This question tests the Multiplication Rule for Probability, which calculates the joint probability of two events. The 'and' here means both cards are hearts in sequence without replacement, so it's conditional: first draw a heart, then another given the first was a heart. The first event A is drawing a heart first, with P(A) = 13/52. The second event B is drawing another heart given the first was a heart, with P(B|A) = 12/51 since one heart is already drawn. Thus, P(A ∩ B) = (13/52) × (12/51) = 1/17, representing the probability of drawing two hearts in a row from the deck. A common mistake is assuming independence and using (13/52) × (13/52), but without replacement, the events are dependent. To avoid this, think of 'and' as 'first draw a heart, then draw another heart,' which highlights the changing deck size in card draws.
A lab follows a two-step procedure. First, a sample is chosen from Batch 1 with probability $P(A)=0.30$ . Then, given it came from Batch 1, the probability it contains contamination is $P(B\mid A)=0.05$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $0.30\times 0.05=0.015$ This question tests the Multiplication Rule for lab sampling procedures. The "both" means we need P(A and B), requiring the sample to be from Batch 1 AND contain contamination. First, a sample is chosen from Batch 1 with probability P(A) = 0.30. Then, given it's from Batch 1, the conditional probability of contamination is P(B|A) = 0.05. By the Multiplication Rule, P(A ∩ B) = P(A) × P(B|A) = 0.30 × 0.05 = 0.015. A common error is using 0.30 × 0.30, assuming the contamination rate equals the batch selection probability, but contamination has its own conditional rate.
A bakery checks cupcakes in two steps. First, a cupcake is chosen at random; $P(A)=0.10$ that it is underbaked. Then, if it is underbaked, the texture test detects this with probability $P(B\mid A)=0.70$ . What is the probability that a cupcake is underbaked and then detected, $P(A\cap B)$ ?
Answer: $0.07$ This question tests the Multiplication Rule for probability. The bakery's checking process involves two sequential events: a cupcake being underbaked and then being detected by the texture test. The first event is that a cupcake is underbaked, with probability P(A) = 0.10. The second event is conditional: given that the cupcake is underbaked, the texture test detects it with probability P(B|A) = 0.70. To find P(A∩B), we multiply: 0.10 × 0.70 = 0.07. A common mistake is to use 0.30 (the complement of 0.70) or to assume independence, but the key is recognizing the conditional nature. Read this as "first the cupcake is underbaked, then it's detected."
A two-step lab test works as follows. First, a sample is contaminated (event $A$ ) with probability $P(A)=0.10$ . Then, given the sample is contaminated, the test returns positive (event $B$ ) with probability $P(B\mid A)=0.90$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $0.09$ This question tests the Multiplication Rule for Probability, calculating the joint probability in a two-step lab test with dependency. The rule sees 'and' as a sequence: first event A, then event B given A. Event A is the sample being contaminated, with P(A) = 0.10. Given contamination, the conditional probability of a positive test (event B) is P(B|A) = 0.90. Multiplying provides P(A ∩ B) = 0.10 × 0.90 = 0.09, the probability of both contamination and positive result. Often, independence is wrongly assumed, leading to adding or squaring, but positivity depends on contamination. Interpret as 'first contaminated, then positive given contamination' for accuracy.
In a sequential game, a player must first solve a puzzle (event $A$ ) and then defeat a boss (event $B$ ). The probability the player solves the puzzle is $P(A)=0.60$ . Given the puzzle is solved, the probability the player defeats the boss is $P(B\mid A)=0.50$ . What is the probability that both $A$ and $B$ occur?
Answer: $0.60\times 0.50=0.30$ This question tests the Multiplication Rule for probability. The "and" indicates a sequential process: first solve the puzzle, then defeat the boss. The first event is solving the puzzle with probability P(A) = 0.60. The second event is defeating the boss given the puzzle was solved, with conditional probability P(B|A) = 0.50. To find P(A and B), we multiply: P(A) × P(B|A) = 0.60 × 0.50 = 0.30. This means only 30% of players complete both challenges. A common mistake is assuming the events are independent and using 0.60 for both probabilities (getting 0.36), but the problem clearly states P(B|A) = 0.50. Always identify the conditional probability when reading "given that."
A mobile app login requires two steps. First, a user enters the correct password with probability $P(A)=0.80$ . Then, given the password is correct, the user passes two-factor authentication with probability $P(B\mid A)=0.95$ . What is the probability that both the password is correct and the user passes two-factor authentication (i.e., $P(A\cap B)$ )?
Answer: $0.80\times 0.95=0.76$ This question tests the Multiplication Rule for Probability, essential for finding the probability of two linked events. The 'and' indicates a sequential process: entering the correct password and then passing authentication, where the second depends on the first. Event A is entering the correct password, with P(A) = 0.80. Event B is passing two-factor authentication given a correct password, with P(B|A) = 0.95. Therefore, P(A ∩ B) = 0.80 × 0.95 = 0.76, the probability of successfully logging in via both steps in the app. A common error is adding the probabilities, assuming independence, but the authentication is conditional on the password. A helpful strategy is to interpret 'and' as 'first correct password, then pass authentication,' focusing on the dependency in login sequences.
A warehouse performs a two-step quality check. First, a package is scanned successfully (event $A$ ). If it is scanned successfully, it is then routed to the correct truck (event $B$ ). If $P(A)=0.90$ and $P(B\mid A)=0.85$ , what is the probability that both $A$ and $B$ occur, $P(A\cap B)$ ?
Answer: $0.765$ This question tests the Multiplication Rule for probability. The "and" indicates sequential events: first the package must be scanned successfully, then it must be routed correctly. The first event A (successful scan) has probability P(A) = 0.90. Given a successful scan, the conditional probability of correct routing is P(B|A) = 0.85. Using the Multiplication Rule, P(A∩B) = P(A) × P(B|A) = 0.90 × 0.85 = 0.765. A common mistake is adding the probabilities (getting 1.75), but remember that "and" means multiply, not add, when events happen in sequence.
A factory inspects products in two steps. First, a product is tested for electrical safety (event $A$ ). If it passes, it is then checked for cosmetic defects (event $B$ ). The probability a product passes the electrical safety test is $P(A)=0.80$ , and the probability it passes the cosmetic check given it passed safety is $P(B\mid A)=0.90$ . What is the probability that both $A$ and $B$ occur (i.e., the product passes both tests)?
Answer: $0.80\times 0.90=0.72$ This question tests the Multiplication Rule for probability. The word "and" here means we need the product to pass the electrical safety test first, then pass the cosmetic check - these events happen sequentially. The first event is passing the electrical safety test with probability P(A) = 0.80. The second event is passing the cosmetic check given the product already passed safety, with conditional probability P(B|A) = 0.90. To find P(A and B), we multiply: P(A) × P(B|A) = 0.80 × 0.90 = 0.72. This means 72% of all products pass both tests. A common mistake is adding the probabilities (getting 1.70), but remember: "and" means multiply when events are conditional, not add.
A game has two sequential events. First, you roll a fair six-sided die; event $A$ is "the roll is a 6," so $P(A)=\frac{1}{6}$ . Then, given that you rolled a 6, you flip a coin that lands heads with probability $P(B\mid A)=\frac{1}{2}$ . What is the probability that both events occur, $P(A\cap B)$ ?
Answer: $\frac{1}{12}$ This question tests the Multiplication Rule for probability. The game has two sequential events: rolling a 6 on the die and then getting heads on the coin flip. The first event is rolling a 6, with probability P(A) = 1/6. The second event is conditional: given that you rolled a 6, the coin lands heads with probability P(B|A) = 1/2. To find P(A∩B), we multiply: (1/6) × (1/2) = 1/12. A common mistake is to add the probabilities or to think the events are independent when they're actually sequential. The strategy is to read "and" as "first roll a 6, then flip heads."

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