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Question 1 / 2
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1
Two students, A and B, select integers from different ranges. Student A selects from 2 to 6 inclusive, and student B from 4 to 11 inclusive. What is the probability they select different numbers?
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Solution: Step 1: Determine the total number of ways A and B can select numbers: - Student A has 5 options (2, 3, 4, 5, 6) - Student B has 8 options (4, 5, 6, 7, 8, 9, 10, 11) Total ways = 5 * 8 = 40 Step 2: Find the number of ways they can select the same number: - Common numbers: 4, 5, 6 Ways to select the same number = 3 Step 3: Calculate the probability of selecting the same number: P(same) = 3/40 Step 4: Calculate the probability of selecting different numbers: P(different) = 1 - P(same) = 1 - 3/40 = (40/40) - (3/40) = 37/40 Therefore, the probability that they select different numbers is 37/40.
2
Three drawers contain different colored balls: one has 6 red and 4 green balls, another has only 5 green balls, and the last one has 4 red balls. A drawer and then a ball are randomly chosen. What is the probability the chosen ball is red?
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Solution: Step 1: Define the probabilities of choosing each drawer: 1/3 for each drawer. Step 2: Calculate the probability of drawing a red ball from each drawer: - Drawer 1: 6 red out of 10 balls = 6/10 = 3/5 - Drawer 2: 0 red balls - Drawer 3: 4 red out of 4 balls = 1 Step 3: Calculate the overall probability of drawing a red ball: P(Red) = P(Drawer 1)*P(Red|Drawer 1) + P(Drawer 2)*P(Red|Drawer 2) + P(Drawer 3)*P(Red|Drawer 3) P(Red) = (1/3)*(3/5) + (1/3)*0 + (1/3)*1 P(Red) = 1/5 + 0 + 1/3 P(Red) = (3/15) + (5/15) P(Red) = 8/15 Therefore, the probability that the chosen ball is red is 8/15.
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