Conditional Probability

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Conditional Probability · Grade 10 · Worksheet 3

1. Charlotte is designing a target for a carnival game. The target is a square board with side length 42 cm. A regular octagon is drawn inside the square such that each vertex of the octagon lies at the midpoint of a side of the square. The region inside the octagon is painted red, and the four corner triangles (outside the octagon but inside the square) are painted blue. If a dart thrown at the board lands at a random point on the board, what is the conditional probability that the dart lands in the red octagon, given that it lands in the bottom-left corner triangle? Express your answer as a simplified fraction. Answer: 0 Solution: The square has side length 42 cm. Each side is divided into two equal halves by the midpoints where the octagon's vertices lie.
   Full step-by-step solution

   Step 1: The square has side length 42 cm. Each side is divided into two equal halves by the midpoints where the octagon's vertices lie. So each corner triangle is a right isosceles triangle with legs of length 21 cm (half the side length). The octagon occupies the central region, and the four corner triangles are outside the octagon. Step 2: The octagon and any corner triangle do not overlap. They are disjoint regions. Therefore, if the dart lands in the bottom-left corner triangle, it cannot possibly also land in the red octagon. Step 3: The conditional probability P(red octagon | bottom-left triangle) = P(red octagon AND bottom-left triangle) / P(bottom-left triangle). Since the intersection of the two regions is empty, P(red octagon AND bottom-left triangle) = 0. Step 4: Therefore, the conditional probability is 0 / P(bottom-left triangle) = 0. The answer is 0.

2. P(A|B) = P(A∩B) / P(B) where P(A) = 0.6, P(B) = 0.8, P(A∩B) = 0.48 = ? Answer: 0.6 Solution: P(A) = 0.6 P(B) = 0.8 P(A ∩ B) = 0.48 We want to find P(A|B).
   Full step-by-step solution

   We are given: P(A) = 0.6 P(B) = 0.8 P(A ∩ B) = 0.48 We want to find P(A|B). Step 1: Recall the formula for conditional probability: P(A|B) = P(A ∩ B) / P(B) Step 2: Substitute the given numbers into the formula: P(A|B) = 0.48 / 0.8 Step 3: Perform the division: 0.48 ÷ 0.8 = 48/80 Step 4: Simplify 48/80: Divide numerator and denominator by 16: 48 ÷ 16 = 3 80 ÷ 16 = 5 So 48/80 = 3/5 Step 5: Convert 3/5 to decimal: 3/5 = 0.6 Step 6: Conclusion: P(A|B) = 0.6 This matches the given correct answer.

3. In a high school, 60% of students play a sport and 40% play a musical instrument. If 25% of students play both a sport and a musical instrument, what is the probability that a randomly selected student plays a sport given that they play a musical instrument? Answer: 0.625 Solution: - S = student plays a sport - M = student plays a musical instrument P(S) = 60% = 0.60 P(M) = 40% = 0.40 P(S and M) = 25% = 0.25 P(S | M) = probability that a student plays a sport given that they play a musical instrument.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define events** Let: - S = student plays a sport - M = student plays a musical instrument Given: P(S) = 60% = 0.60 P(M) = 40% = 0.40 P(S and M) = 25% = 0.25 --- **Step 2: Identify what is being asked** We want: P(S | M) = probability that a student plays a sport given that they play a musical instrument. --- **Step 3: Recall the conditional probability formula** P(S | M) = P(S and M) / P(M) --- **Step 4: Substitute the given numbers** P(S | M) = 0.25 / 0.40 --- **Step 5: Perform the division** 0.25 / 0.40 = 25/40 = 5/8 --- **Step 6: Convert to decimal** 5/8 = 0.625 --- **Step 7: Interpret the result** The probability that a randomly selected student plays a sport, given that they play a musical instrument, is 0.625. --- **Final Answer:** 0.625

4. P(A|B) = P(A∩B) / P(B) where P(A) = 0.9, P(B) = 0.8, P(A∩B) = 0.72 = ? Answer: 0.9 Solution: Write the conditional probability formula: P(A|B) = P(A∩B) / P(B) Substitute the given values: P(A|B) = 0.72 / 0.8 Perform the division: 0.72 ÷ 0.8 = 0.9 The conditional probability P(A|B) is 0.9
   Full step-by-step solution

   Step 1: Write the conditional probability formula: P(A|B) = P(A∩B) / P(B) Step 2: Substitute the given values: P(A|B) = 0.72 / 0.8 Step 3: Perform the division: 0.72 ÷ 0.8 = 0.9 Step 4: The conditional probability P(A|B) is 0.9

5. Aroha runs a diagnostic clinic for a rare plant disease. The test for the disease is 95% accurate at detecting the disease when it is present (true positive rate), but it also has a 4% false positive rate, meaning it incorrectly indicates the disease in 4% of healthy plants. Only 2% of the plants in the region actually have the disease. If a randomly selected plant tests positive for the disease, what is the probability that it actually has the disease? Express your answer as a percentage rounded to two decimal places. Answer: 32.65% Solution: Define events. Let D be the event that a plant has the disease, and T+ be the event that a plant tests positive. P(D) = 0.02 (2% have disease) P(not D) = 0.98 (98% are healthy) P(T+ | D) = 0.95 (true positive rate) P(T+ | not D) = 0.04 (false positive rate) We want P(D | T+), the probability a…
   Full step-by-step solution

   Step 1: Define events. Let D be the event that a plant has the disease, and T+ be the event that a plant tests positive. Step 2: Given probabilities: P(D) = 0.02 (2% have disease) P(not D) = 0.98 (98% are healthy) P(T+ | D) = 0.95 (true positive rate) P(T+ | not D) = 0.04 (false positive rate) Step 3: We want P(D | T+), the probability a plant has the disease given it tested positive. Step 4: Use Bayes' theorem: P(D | T+) = [P(T+ | D) * P(D)] / [P(T+ | D) * P(D) + P(T+ | not D) * P(not D)] Step 5: Calculate the numerator: 0.95 * 0.02 = 0.019 Step 6: Calculate the denominator: (0.95 * 0.02) + (0.04 * 0.98) = 0.019 + 0.0392 = 0.0582 Step 7: P(D | T+) = 0.019 / 0.0582 ≈ 0.32646 Step 8: Convert to percentage: 0.32646 * 100 ≈ 32.65% The answer is 32.65%.