Question 16: a) Define P(A ∩ B) for independent events. b) Two cards are drawn without replacement. Find the probability both are face cards. c) Draw a tree diagram for face card probabilities.

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Asked on April 4, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-04T09:06:57.351Z

Here's the solution to question 16: a) If two events A and B are independent events, the formula for finding P(A B) is: P(A B) = P(A) × P(B) The final answer is P(A B) = P(A) × P(B). b) Two cards are drawn randomly one after another without replacement. Total number of cards = 52. Number of face cards (King, Queen, Jack) = 3 cards/suit × 4 suits = 12 face cards. Step 1: Calculate the probability of the first card being a face card. P(1st Face) = Number of face cardsTotal cards = (12)/(52) = (3)/(13) Step 2: Calculate the probability of the second card being a face card, given the first was a face card and not replaced. After drawing one face card, there are 51 cards remaining, and 11 face cards remaining. P(2nd Face | 1st Face) = Remaining face cardsRemaining cards = (11)/(51) Step 3: Calculate the probability of both cards being face cards. P(Both Face) = P(1st Face) × P(2nd Face | 1st Face) P(Both Face) = (12)/(52) × (11)/(51) = (3)/(13) × (11)/(51) = (33)/(663) = (11)/(221) The final answer is (11)/(221). c) Show the probability of all the possible outcomes of getting or not-getting a face card in a tree diagram (without replacement). Let F be the event of getting a face card. Let F' be the event of not getting a face card. Number of face cards = 12. Number of non-face cards = 52 - 12 = 40. Tree Diagram: First Draw: Probability of getting a face card (F_1): P(F_1) = (12)/(52) = (3)/(13) Probability of not getting a face card (F'_1): P(F'_1) = (40)/(52) = (10)/(13) Second Draw (without replacement): If the first card was a face card (F_1): Remaining cards = 51 Remaining face cards = 11 Remaining non-face cards = 40 Probability of getting a face card (F_2 | F_1): P(F_2 | F_1) = (11)/(51) Probability of not getting a face card (F'_2 | F_1): P(F'_2 | F_1) = (40)/(51) If the first card was not a face card (F'_1): Remaining cards = 51 Remaining face cards = 12 Remaining non-face cards = 39 Probability of getting a face card (F_2 | F'_1): P(F_2 | F'_1) = (12)/(51) = (4)/(17) Probability of not getting a face card (F'_2 | F'_1): P(F'_2 | F'_1) = (39)/(51) = (13)/(17) Possible Outcomes and their Probabilities: 1. Both Face Cards (F_1 and F_2): P(F_1 F_2) = P(F_1) × P(F_2 | F_1) = (12)/(52) × (11)/(51) = (3)/(13) × (11)/(51) = (11)/(221) 2. First Face, Second Not Face (F_1 and F'_2): P(F_1 F'_2) = P(F_1) × P(F'_2 | F_1) = (12)/(52) × (40)/(51) = (3)/(13) × (40)/(51) = (40)/(221) 3. First Not Face, Second Face (F'_1 and F_2): P(F'_1 F_2) = P(F'_1) × P(F_2 | F'_1) = (40)/(52) × (12)/(51) = (10)/(13) × (12)/(51) = (40)/(221) 4. Both Not Face Cards (F'_1 and F'_2): P(F'_1 F'_2) = P(F'_1) × P(F'_2 | F'_1) = (40)/(52) × (39)/(51) = (10)/(13) × (39)/(51) = (130)/(221) The tree diagram visually represents these branches and their associated probabilities. d) If two cards are drawn randomly one after another with replacement. Step 1: Calculate the probability that both cards are face cards (with replacement). Number of face cards = 12. P(Both Face, with replacement) = P(1st Face) × P(2nd Face) P(Both Face, with replacement) = (12)/(52) × (12)/(52) = (3)/(13) × (3)/(13) = (9)/(169) Step 2: Calculate the probability that both cards are ace (with replacement). Number of ace cards = 4. P(Both Ace, with replacement) = P(1st Ace) × P(2nd Ace) P(Both Ace, with replacement) = (4)/(52) × (4)/(52) = (1)/(13) × (1)/(13) = (1)/(169) Step 3: Determine how many times more is the probability that both are face cards than the probability that both cards are ace. Ratio = P(Both Face, with replacement)P(

Accepted Answer · 2026-04-04T09:06:57.351Z

Here's the solution to question 16: a) If two events A and B are independent events, the formula for finding P(A B) is: P(A B) = P(A) × P(B) The final answer is P(A B) = P(A) × P(B). b) Two cards are drawn randomly one after another without replacement. Total number of cards = 52. Number of face cards (King, Queen, Jack) = 3 cards/suit × 4 suits = 12 face cards. Step 1: Calculate the probability of the first card being a face card. P(1st Face) = Number of face cardsTotal cards = (12)/(52) = (3)/(13) Step 2: Calculate the probability of the second card being a face card, given the first was a face card and not replaced. After drawing one face card, there are 51 cards remaining, and 11 face cards remaining. P(2nd Face | 1st Face) = Remaining face cardsRemaining cards = (11)/(51) Step 3: Calculate the probability of both cards being face cards. P(Both Face) = P(1st Face) × P(2nd Face | 1st Face) P(Both Face) = (12)/(52) × (11)/(51) = (3)/(13) × (11)/(51) = (33)/(663) = (11)/(221) The final answer is (11)/(221). c) Show the probability of all the possible outcomes of getting or not-getting a face card in a tree diagram (without replacement). Let F be the event of getting a face card. Let F' be the event of not getting a face card. Number of face cards = 12. Number of non-face cards = 52 - 12 = 40. Tree Diagram: First Draw: Probability of getting a face card (F_1): P(F_1) = (12)/(52) = (3)/(13) Probability of not getting a face card (F'_1): P(F'_1) = (40)/(52) = (10)/(13) Second Draw (without replacement): If the first card was a face card (F_1): Remaining cards = 51 Remaining face cards = 11 Remaining non-face cards = 40 Probability of getting a face card (F_2 | F_1): P(F_2 | F_1) = (11)/(51) Probability of not getting a face card (F'_2 | F_1): P(F'_2 | F_1) = (40)/(51) If the first card was not a face card (F'_1): Remaining cards = 51 Remaining face cards = 12 Remaining non-face cards = 39 Probability of getting a face card (F_2 | F'_1): P(F_2 | F'_1) = (12)/(51) = (4)/(17) Probability of not getting a face card (F'_2 | F'_1): P(F'_2 | F'_1) = (39)/(51) = (13)/(17) Possible Outcomes and their Probabilities: 1. Both Face Cards (F_1 and F_2): P(F_1 F_2) = P(F_1) × P(F_2 | F_1) = (12)/(52) × (11)/(51) = (3)/(13) × (11)/(51) = (11)/(221) 2. First Face, Second Not Face (F_1 and F'_2): P(F_1 F'_2) = P(F_1) × P(F'_2 | F_1) = (12)/(52) × (40)/(51) = (3)/(13) × (40)/(51) = (40)/(221) 3. First Not Face, Second Face (F'_1 and F_2): P(F'_1 F_2) = P(F'_1) × P(F_2 | F'_1) = (40)/(52) × (12)/(51) = (10)/(13) × (12)/(51) = (40)/(221) 4. Both Not Face Cards (F'_1 and F'_2): P(F'_1 F'_2) = P(F'_1) × P(F'_2 | F'_1) = (40)/(52) × (39)/(51) = (10)/(13) × (39)/(51) = (130)/(221) The tree diagram visually represents these branches and their associated probabilities. d) If two cards are drawn randomly one after another with replacement. Step 1: Calculate the probability that both cards are face cards (with replacement). Number of face cards = 12. P(Both Face, with replacement) = P(1st Face) × P(2nd Face) P(Both Face, with replacement) = (12)/(52) × (12)/(52) = (3)/(13) × (3)/(13) = (9)/(169) Step 2: Calculate the probability that both cards are ace (with replacement). Number of ace cards = 4. P(Both Ace, with replacement) = P(1st Ace) × P(2nd Ace) P(Both Ace, with replacement) = (4)/(52) × (4)/(52) = (1)/(13) × (1)/(13) = (1)/(169) Step 3: Determine how many times more is the probability that both are face cards than the probability that both cards are ace. Ratio = P(Both Face, with replacement)P(

Question 16: a) Define P(A ∩ B) for independent events. b) Two cards are drawn without replacement. Find the probability both are face cards. c) Draw a tree diagram for face card probabilities.

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Question 16: a) Define P(A ∩ B) for independent events. b) Two cards are drawn without replacement. Find the probability both are face cards. c) Draw a tree diagram for face card probabilities.

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