This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here are the solutions to the probability questions: There are 13 cubes numbered from 1 to 13. The odd numbers are \1, 3, 5, 7, 9, 11, 13\, so there are 7 odd cubes. The even numbers are \2, 4, 6, 8, 10, 12\, so there are 6 even cubes. Two cubes are drawn one after the other without replacement. a) What is the probability that both cubes are odd numbered? Step 1: Find the probability of drawing an odd cube first. There are 7 odd cubes out of 13 total cubes. P(1st cube is odd) = (7)/(13) Step 2: Find the probability of drawing a second odd cube, given the first was odd and not replaced. After drawing one odd cube, there are 6 odd cubes left and a total of 12 cubes remaining. P(2nd cube is odd | 1st cube is odd) = (6)/(12) = (1)/(2) Step 3: Multiply the probabilities to find the probability that both cubes are odd. P(both are odd) = P(1st odd) × P(2nd odd | 1st odd) P(both are odd) = (7)/(13) × (1)/(2) P(both are odd) = (7)/(26) The probability that both cubes are odd numbered is (7)/(26). b) What is the probability that only one is even numbered? This means one cube is even and the other is odd. There are two possible scenarios: Scenario 1: The first cube is even, and the second cube is odd. Scenario 2: The first cube is odd, and the second cube is even. Step 1: Calculate the probability for Scenario 1 (1st Even, 2nd Odd). P(1st Even) = (6)/(13) After drawing one even cube, there are 7 odd cubes left and 12 total cubes remaining. P(2nd Odd | 1st Even) = (7)/(12) P(1st Even, 2nd Odd) = (6)/(13) × (7)/(12) = (1)/(13) × (7)/(2) = (7)/(26) Step 2: Calculate the probability for Scenario 2 (1st Odd, 2nd Even). P(1st Odd) = (7)/(13) After drawing one odd cube, there are 6 even cubes left and 12 total cubes remaining. P(2nd Even | 1st Odd) = (6)/(12) = (1)/(2) P(1st Odd, 2nd Even) = (7)/(13) × (1)/(2) = (7)/(26) Step 3: Add the probabilities of the two scenarios. P(only one is even) = P(1st Even, 2nd Odd) + P(1st Odd, 2nd Even) P(only one is even) = (7)/(26) + (7)/(26) P(only one is even) = (14)/(26) = (7)/(13) The probability that only one cube is even numbered is (7)/(13). 3 done, 2 left today. You're making progress.