Probability Rules

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Probability Rules · Grade 10 · Worksheet 3

1. P(Emma passes math) = 0.7, P(Emma passes science | Emma passes math) = 0.9, find P(Emma passes math and science) Answer: 0.63 Solution: We are given P(Emma passes math) = 0.7 and P(Emma passes science | Emma passes math) = 0.9 The multiplication rule states: P(A and B) = P(A) × P(B|A) Let A = Emma passes math, B = Emma passes science P(A and B) = P(A) × P(B|A) = 0.7 × 0.9 0.7 × 0.9 = 0.63 Therefore, P(Emma passes math and…
   Full step-by-step solution

   Step 1: We are given P(Emma passes math) = 0.7 and P(Emma passes science | Emma passes math) = 0.9 Step 2: The multiplication rule states: P(A and B) = P(A) × P(B|A) Step 3: Let A = Emma passes math, B = Emma passes science Step 4: P(A and B) = P(A) × P(B|A) = 0.7 × 0.9 Step 5: 0.7 × 0.9 = 0.63 Step 6: Therefore, P(Emma passes math and science) = 0.63

2. P(Ava) = 0.87, P(Noah) = 0.65, P(Ava and Noah) = 0.5655. Find P(Ava|Noah). Answer: 0.87 Solution: Write the conditional probability formula: P(Ava|Noah) = P(Ava and Noah) / P(Noah) Substitute the given values: P(Ava|Noah) = 0.5655 / 0.65 Perform the division: 0.5655 ÷ 0.65 = 0.87 The conditional probability is 0.87
   Full step-by-step solution

   Step 1: Write the conditional probability formula: P(Ava|Noah) = P(Ava and Noah) / P(Noah) Step 2: Substitute the given values: P(Ava|Noah) = 0.5655 / 0.65 Step 3: Perform the division: 0.5655 ÷ 0.65 = 0.87 Step 4: The conditional probability is 0.87

3. Mason is analyzing the accuracy of a new medical test for detecting a rare virus. The probability of having the virus is 0.08. If a person has the virus, the test correctly identifies it 96% of the time. If a person does not have the virus, the test incorrectly shows a positive result 11% of the time. What is the probability that a person who tests positive actually has the virus? Answer: A. 0.44 Solution: P(V) = 0.08 P(+|V) = 0.96 P(+|not V) = 0.11 P(not V) = 1 - P(V) = 1 - 0.08 = 0.92 P(+) = P(+|V) × P(V) + P(+|not V) × P(not V) P(+) = (0.96 × 0.08) + (0.11 × 0.92) P(+) = 0.0768 + 0.1012 P(+) = 0.178 P(V|+) = [P(+|V) × P(V)] / P(+) P(V|+) = (0.96 × 0.08) / 0.178 P(V|+) = 0.0768 / 0.178 P(V|+) ≈…
   Full step-by-step solution

   Step 1: Define the events Let V be the event that a person has the virus Let + be the event that the test is positive Step 2: Identify the given probabilities P(V) = 0.08 P(+|V) = 0.96 P(+|not V) = 0.11 Step 3: Calculate P(not V) P(not V) = 1 - P(V) = 1 - 0.08 = 0.92 Step 4: Calculate P(+) using the law of total probability P(+) = P(+|V) × P(V) + P(+|not V) × P(not V) P(+) = (0.96 × 0.08) + (0.11 × 0.92) P(+) = 0.0768 + 0.1012 P(+) = 0.178 Step 5: Apply Bayes' theorem to find P(V|+) P(V|+) = [P(+|V) × P(V)] / P(+) P(V|+) = (0.96 × 0.08) / 0.178 P(V|+) = 0.0768 / 0.178 P(V|+) ≈ 0.43146 ≈ 0.44 The correct answer is 0.44.

4. P(Aroha) = 0.85, P(Tane|Aroha) = 0.72, P(Aroha and Tane) = ? Answer: 0.612 Solution: Recall the multiplication rule for conditional probability: P(A and B) = P(A) × P(B|A). Substitute the given values: P(Aroha and Tane) = P(Aroha) × P(Tane|Aroha). Calculate: 0.85 × 0.72 = 0.612.
   Full step-by-step solution

   Step 1: Recall the multiplication rule for conditional probability: P(A and B) = P(A) × P(B|A). Step 2: Substitute the given values: P(Aroha and Tane) = P(Aroha) × P(Tane|Aroha). Step 3: Calculate: 0.85 × 0.72 = 0.612. Step 4: The probability that both events occur is 0.612.

5. P(A) = 0.85, P(B) = 0.72, P(A and B) = 0.612. Find P(A|B). Answer: 0.85 Solution: Write the conditional probability formula: P(A|B) = P(A and B) / P(B) Substitute the given values: P(A|B) = 0.612 / 0.72 Calculate the division: 0.612 ÷ 0.72 = 0.85 The conditional probability P(A|B) is 0.85
   Full step-by-step solution

   Step 1: Write the conditional probability formula: P(A|B) = P(A and B) / P(B) Step 2: Substitute the given values: P(A|B) = 0.612 / 0.72 Step 3: Calculate the division: 0.612 ÷ 0.72 = 0.85 Step 4: The conditional probability P(A|B) is 0.85

6. Matiu is analyzing a geometric probability problem involving two overlapping squares. Square A has side length 12 cm and Square B has side length 15 cm. The squares overlap such that the area of their intersection is 45 square cm. If a point is randomly chosen inside Square B, what is the probability that it also lies inside Square A? Answer: 0.2 Solution: Identify the relevant probability formula. We need P(A|B) = P(A and B) / P(B). Calculate P(A and B), which is the probability a point is in both squares.
   Full step-by-step solution

   Step 1: Identify the relevant probability formula. We need P(A|B) = P(A and B) / P(B). Step 2: Calculate P(A and B), which is the probability a point is in both squares. This equals the intersection area divided by the total area of Square B: 45 / (15 × 15) = 45 / 225 = 0.2. Step 3: Since we're given that the point is already in Square B, P(B) = 1. Step 4: Apply the conditional probability formula: P(A|B) = 0.2 / 1 = 0.2. The answer is 0.2.

7. P(Aroha passes math) = 0.85, P(Aroha passes science) = 0.90, P(Aroha passes both) = 0.80. Find P(Aroha passes science | Aroha passes math). Answer: 0.941 Solution: Identify the events. Let A = Aroha passes science, B = Aroha passes math. Write down the given probabilities: P(B) = 0.85, P(A) = 0.90, P(A and B) = 0.80.
   Full step-by-step solution

   Step 1: Identify the events. Let A = Aroha passes science, B = Aroha passes math. Step 2: Write down the given probabilities: P(B) = 0.85, P(A) = 0.90, P(A and B) = 0.80. Step 3: Apply the conditional probability formula: P(A|B) = P(A and B) / P(B). Step 4: Substitute the values: P(A|B) = 0.80 / 0.85. Step 5: Calculate the division: 0.80 / 0.85 = 80/85 = 16/17 ≈ 0.941176. Step 6: Round to three decimal places: 0.941. The answer is 0.941.