Conditional Probability

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Conditional Probability · Grade 8 · Worksheet 2

1. Tane has a rectangular board divided into a grid of 10 rows and 10 columns, making 100 equal squares. The squares are numbered 1 to 100 row by row, starting from the top left. He randomly selects one square. Let event A be selecting a square with a number that is a multiple of 11. Let event B be selecting a square in the first 4 rows (rows 1–4). Are events A and B independent? Show your reasoning by calculating P(A and B) and comparing it to P(A) * P(B). Answer: No, they are not independent. Solution: Total squares = 100. Event A: numbers that are multiples of 11 from 1 to 100: 11, 22, 33, 44, 55, 66, 77, 88, 99. That's 9 numbers.
   Full step-by-step solution

   Step 1: Total squares = 100. Event A: numbers that are multiples of 11 from 1 to 100: 11, 22, 33, 44, 55, 66, 77, 88, 99. That's 9 numbers. So P(A) = 9/100. Step 2: Event B: squares in rows 1–4. Each row has 10 squares, so rows 1–4 have 4 × 10 = 40 squares. So P(B) = 40/100 = 2/5. Step 3: Event A and B: multiples of 11 that are in rows 1–4 (numbers 1–40). From the list: 11, 22, 33 are in rows 1–4. That's 3 numbers. So P(A and B) = 3/100. Step 4: Check independence: If independent, P(A and B) = P(A) * P(B). Compute P(A) * P(B) = (9/100) * (40/100) = 360/10000 = 9/250 = 0.036. But P(A and B) = 3/100 = 0.03. Since 0.03 ≠ 0.036, the events are not independent. The answer is: No, they are not independent.

2. A school survey found that 60% of students play sports and 40% of students play a musical instrument. If 24% of students do both activities, are playing sports and playing a musical instrument independent events? Show your reasoning. Answer: B. no Solution: Two events are independent if the probability of both occurring equals the product of their individual probabilities. This means one event happening doesn't affect the likelihood of the other event happening.
   Full step-by-step solution

   Two events are independent if the probability of both occurring equals the product of their individual probabilities. This means one event happening doesn't affect the likelihood of the other event happening. You can test for independence using this multiplication rule.

3. A bag contains 3 red marbles, 5 blue marbles, and 2 green marbles. If two marbles are drawn at random without replacement, what is the probability that both marbles are red? Answer: 1/15 Solution: Calculate the total number of marbles. 3 red + 5 blue + 2 green = 10 marbles. Calculate the probability that the first marble drawn is red.
   Full step-by-step solution

   Step 1: Calculate the total number of marbles. 3 red + 5 blue + 2 green = 10 marbles. Step 2: Calculate the probability that the first marble drawn is red. P(First Red) = Number of red marbles / Total marbles = 3/10. Step 3: If the first marble is red, there are now 2 red marbles left and 9 total marbles remaining. Step 4: Calculate the probability that the second marble is red given the first was red. P(Second Red | First Red) = 2/9. Step 5: Multiply the probabilities to find the probability both events occur. P(Both Red) = (3/10) × (2/9) = 6/90. Step 6: Simplify the fraction 6/90 by dividing the numerator and denominator by 6. 6/90 = 1/15. The final answer is 1/15.

4. Mere has a rectangular board divided into a grid of 4 rows and 6 columns. Each cell is colored either red or blue. The board is shown as follows: The first row has all 6 cells blue. The second row has the first 4 cells blue and the last 2 cells red. The third row has the first 2 cells red and the last 4 cells blue. The fourth row has all 6 cells red. A cell is chosen at random. Let event A be that the cell is in an even-numbered column (columns numbered 1 to 6 from left to right). Let event B be that the cell is blue. Are events A and B independent? Justify your answer with probability calculations. Answer: No, they are not independent Solution: Count total cells. The board has 4 rows and 6 columns, so total cells = 4 × 6 = 24. Count blue cells.
   Full step-by-step solution

   Step 1: Count total cells. The board has 4 rows and 6 columns, so total cells = 4 × 6 = 24. Step 2: Count blue cells. Row 1: 6 blue. Row 2: 4 blue. Row 3: 4 blue. Row 4: 0 blue. Total blue = 6 + 4 + 4 + 0 = 14. So P(B) = 14/24 = 7/12. Step 3: Count cells in even-numbered columns. Columns 2, 4, 6 are even. Each column has 4 cells (one per row). So total cells in even columns = 3 columns × 4 cells = 12. So P(A) = 12/24 = 1/2. Step 4: Count cells that are both blue AND in an even column. Even columns are 2, 4, 6. For each row, count blue cells in those columns: - Row 1: all 6 blue, so in columns 2,4,6 all blue → 3 blue cells. - Row 2: first 4 blue (columns 1-4), so columns 2 and 4 are blue, column 6 is red → 2 blue cells. - Row 3: last 4 blue (columns 3-6), so columns 4 and 6 are blue, column 2 is red → 2 blue cells. - Row 4: all red, so 0 blue cells. Total blue in even columns = 3 + 2 + 2 + 0 = 7. So P(A and B) = 7/24. Step 5: Check independence. Compute P(A) × P(B) = (1/2) × (7/12) = 7/24. Step 6: Compare. P(A and B) = 7/24 equals P(A) × P(B) = 7/24. So the events are independent. The answer is: Yes, they are independent.

5. A school survey found that 60% of students play sports and 25% of students play a musical instrument. If 15% of students do both activities, what percentage of students play sports but do not play a musical instrument? Answer: 45 Solution: - Percentage of students who play sports = 60% - Percentage of students who play a musical instrument = 25% - Percentage of students who do both = 15% We want: percentage of students who play sports but do not play a musical instrument.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the given data** - Percentage of students who play sports = 60% - Percentage of students who play a musical instrument = 25% - Percentage of students who do both = 15% --- **Step 2: Interpret the question** We want: percentage of students who play sports but do **not** play a musical instrument. That means: From the sports players, remove those who also play an instrument. --- **Step 3: Use a set reasoning approach** Let S = set of students who play sports Let M = set of students who play a musical instrument We know: P(S) = 60% P(M) = 25% P(S and M) = 15% We want: P(S and not M) = ? --- **Step 4: Apply the formula** P(S and not M) = P(S) - P(S and M) = 60% - 15% = 45% --- **Step 5: Conclusion** 45% of students play sports but do not play a musical instrument. --- **Final answer:** 45

6. A local bookstore is analyzing reading habits. They found that 40% of their customers buy fiction books and 30% buy non-fiction books. If buying fiction and non-fiction books are independent events, what percentage of customers would you expect to buy both types of books? Answer: 12 Solution: - Probability of buying fiction books = 40% = 0.40 - Probability of buying non-fiction books = 30% = 0.30 Since the events are independent, multiply the probabilities - Probability of buying both = 0.40 × 0.30 - 0.40 × 0.30 = 0.12 - 0.12 = 12% The answer is 12.
   Full step-by-step solution

   Step 1: Identify the given probabilities - Probability of buying fiction books = 40% = 0.40 - Probability of buying non-fiction books = 30% = 0.30 Step 2: Since the events are independent, multiply the probabilities - Probability of buying both = 0.40 × 0.30 Step 3: Calculate the product - 0.40 × 0.30 = 0.12 Step 4: Convert to percentage - 0.12 = 12% The answer is 12.

7. Sophia rolls a fair 8-sided die numbered 1 through 8. Event A is rolling a prime number. Event B is rolling a number greater than 5. Are events A and B independent? Show your work using the definition of independence. Answer: No, they are not independent. Solution: List outcomes. Sample space S = {1,2,3,4,5,6,7,8}. Event A (prime numbers): {2,3,5,7}.
   Full step-by-step solution

   Step 1: List outcomes. Sample space S = {1,2,3,4,5,6,7,8}. Event A (prime numbers): {2,3,5,7}. Event B (numbers > 5): {6,7,8}. A and B = {7}. Step 2: Calculate probabilities. P(A) = 4/8 = 1/2. P(B) = 3/8. P(A and B) = 1/8. Step 3: Check independence. P(A) × P(B) = (1/2) × (3/8) = 3/16. P(A and B) = 1/8 = 2/16. Step 4: Since 3/16 ≠ 2/16, P(A) × P(B) ≠ P(A and B). Therefore, events A and B are not independent. The answer is: No, they are not independent.