Probability Rules

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Probability Rules · Grade 10 · Worksheet 2

1. Matiu is analyzing a geometric pattern of colored tiles arranged in a rectangular grid. The grid has 15 rows and 12 columns. Each tile is either blue or yellow. Matiu randomly selects one tile from the grid. Given that the selected tile is from an even-numbered row, what is the probability that it is also from an odd-numbered column? Answer: 0.5 Solution: The grid has 15 rows and 12 columns, so there are 15 × 12 = 180 total tiles. Even-numbered rows are rows 2, 4, 6, 8, 10, 12, 14. That's 7 rows.
   Full step-by-step solution

   Step 1: The grid has 15 rows and 12 columns, so there are 15 × 12 = 180 total tiles. Step 2: Even-numbered rows are rows 2, 4, 6, 8, 10, 12, 14. That's 7 rows. Step 3: Each even-numbered row has 12 tiles, so there are 7 × 12 = 84 tiles in even-numbered rows. Step 4: Odd-numbered columns are columns 1, 3, 5, 7, 9, 11. That's 6 columns. Step 5: For each even-numbered row, there are 6 tiles in odd-numbered columns. Step 6: So the number of tiles in even-numbered rows AND odd-numbered columns is 7 × 6 = 42. Step 7: The probability is P(odd column | even row) = (tiles in even rows and odd columns) / (tiles in even rows) = 42 / 84 = 0.5. The answer is 0.5.

2. P(disease) = 0.03, P(positive|disease) = 0.97, P(positive|no disease) = 0.07. Find P(disease|positive). Answer: 0.3 Solution: Calculate P(no disease) = 1 - P(disease) = 1 - 0.03 = 0.97 Calculate P(positive and disease) = P(positive|disease) × P(disease) = 0.97 × 0.03 = 0.0291 Calculate P(positive and no disease) = P(positive|no disease) × P(no disease) = 0.07 × 0.97 = 0.0679 Calculate P(positive) = P(positive and…
   Full step-by-step solution

   Step 1: Calculate P(no disease) = 1 - P(disease) = 1 - 0.03 = 0.97 Step 2: Calculate P(positive and disease) = P(positive|disease) × P(disease) = 0.97 × 0.03 = 0.0291 Step 3: Calculate P(positive and no disease) = P(positive|no disease) × P(no disease) = 0.07 × 0.97 = 0.0679 Step 4: Calculate P(positive) = P(positive and disease) + P(positive and no disease) = 0.0291 + 0.0679 = 0.097 Step 5: Apply Bayes' theorem: P(disease|positive) = P(positive and disease) / P(positive) = 0.0291 / 0.097 = 0.3 The answer is 0.3.

3. Tane is testing a new fertilizer on his tomato plants. The probability that a plant grows unusually large tomatoes with the fertilizer is 0.7. The probability that a plant both grows unusually large tomatoes and has increased disease resistance is 0.35. Given that a plant grew unusually large tomatoes, what is the probability it also has increased disease resistance? Answer: 0.5 Solution: P(large tomatoes) = 0.7 P(large tomatoes AND disease resistance) = 0.35 P(disease resistance | large tomatoes) = P(large tomatoes AND disease resistance) / P(large tomatoes) P(disease resistance | large tomatoes) = 0.35 / 0.7 0.35 / 0.7 = 0.5 The answer is 0.5.
   Full step-by-step solution

   Step 1: Identify the given probabilities P(large tomatoes) = 0.7 P(large tomatoes AND disease resistance) = 0.35 Step 2: Apply the conditional probability formula P(disease resistance | large tomatoes) = P(large tomatoes AND disease resistance) / P(large tomatoes) Step 3: Substitute the values P(disease resistance | large tomatoes) = 0.35 / 0.7 Step 4: Calculate the result 0.35 / 0.7 = 0.5 The answer is 0.5.

4. P(Liam) = 0.75, P(Olivia) = 0.45, P(Liam and Olivia) = 0.3375. Find P(Liam|Olivia). Answer: 0.75 Solution: P(Liam) = 0.75 P(Olivia) = 0.45 P(Liam and Olivia) = 0.3375 P(Liam|Olivia) = P(Liam and Olivia) / P(Olivia) P(Liam|Olivia) = 0.3375 / 0.45 0.3375 ÷ 0.45 = 0.75 Since P(Liam|Olivia) = P(Liam) = 0.75, the events are independent.
   Full step-by-step solution

   Step 1: Identify the given probabilities P(Liam) = 0.75 P(Olivia) = 0.45 P(Liam and Olivia) = 0.3375 Step 2: Apply the conditional probability formula P(Liam|Olivia) = P(Liam and Olivia) / P(Olivia) Step 3: Substitute the values P(Liam|Olivia) = 0.3375 / 0.45 Step 4: Calculate the result 0.3375 ÷ 0.45 = 0.75 Step 5: Verify the result Since P(Liam|Olivia) = P(Liam) = 0.75, the events are independent. The answer is 0.75.

5. P(Aroha passes math) = 0.92, P(Aroha passes science) = 0.88, P(Aroha passes both) = 0.81. Find P(Aroha passes science | Aroha passes math). Answer: 0.8804 Solution: Identify the events. Let A = Aroha passes science, B = Aroha passes math.
   Full step-by-step solution

   Step 1: Identify the events. Let A = Aroha passes science, B = Aroha passes math. Step 2: Write the conditional probability formula: P(A|B) = P(A and B) / P(B) Step 3: Substitute the given values: P(A and B) = 0.81, P(B) = 0.92 Step 4: Calculate: P(A|B) = 0.81 / 0.92 Step 5: Perform the division: 0.81 ÷ 0.92 = 0.8804347826 Step 6: Round to 4 decimal places: 0.8804 Step 7: The answer is 0.8804

6. P(Matiu passes test) = 0.4, P(Matiu studies | Matiu passes test) = 0.8. Find P(Matiu studies and passes test). Answer: 0.32 Solution: P(passes test) = 0.4 P(studies | passes test) = 0.8 Apply the multiplication rule for conditional probability P(studies and passes test) = P(passes test) × P(studies | passes test) P(studies and passes test) = 0.4 × 0.8 P(studies and passes test) = 0.32 The answer is 0.32.
   Full step-by-step solution

   Step 1: Identify the given probabilities P(passes test) = 0.4 P(studies | passes test) = 0.8 Step 2: Apply the multiplication rule for conditional probability P(studies and passes test) = P(passes test) × P(studies | passes test) Step 3: Substitute the given values P(studies and passes test) = 0.4 × 0.8 Step 4: Calculate the result P(studies and passes test) = 0.32 The answer is 0.32.

7. Mere is analyzing quality control data for a factory that produces electronic components. The factory has two production lines: Line X produces 65% of the components, while Line Y produces 35%. Historical data shows that 12% of components from Line X are defective, while 18% from Line Y are defective. If a randomly selected component is found to be defective, what is the probability it came from Line Y? Answer: A. 0.447 Solution: Let Y be the event that a component comes from Line Y P(Y) = 0.35 (Line Y produces 35% of components) P(X) = 0.65 (Line X produces 65% of components) P(D|X) = 0.12 (12% of Line X components are defective) P(D|Y) = 0.18 (18% of Line Y components are defective) P(D) = P(D|X)P(X) + P(D|Y)P(Y) P(D)…
   Full step-by-step solution

   Step 1: Define the events Let Y be the event that a component comes from Line Y Let D be the event that a component is defective Step 2: Write down the given probabilities P(Y) = 0.35 (Line Y produces 35% of components) P(X) = 0.65 (Line X produces 65% of components) P(D|X) = 0.12 (12% of Line X components are defective) P(D|Y) = 0.18 (18% of Line Y components are defective) Step 3: Calculate P(D) using the law of total probability P(D) = P(D|X)P(X) + P(D|Y)P(Y) P(D) = (0.12)(0.65) + (0.18)(0.35) P(D) = 0.078 + 0.063 P(D) = 0.141 Step 4: Apply Bayes' theorem to find P(Y|D) P(Y|D) = P(D|Y)P(Y) / P(D) P(Y|D) = (0.18)(0.35) / 0.141 P(Y|D) = 0.063 / 0.141 P(Y|D) = 0.447 The probability that a defective component came from Line Y is 0.447.