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 is the solution to question 11. Part 11 (a): The set of numbers is S = \3, 5, 6, 8, 11, 14, 17\. The total number of elements in the set is n(S) = 7. Step 1: Find the probability of choosing an even number. The even numbers in S are \6, 8, 14\. The number of even numbers is n(even) = 3. The probability of choosing an even number is: P(even) = n(even)n(S) = (3)/(7) The probability of choosing an even number is (3)/(7). Step 2: Find the probability of choosing a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The prime numbers in S are \3, 5, 11, 17\. The number of prime numbers is n(prime) = 4. The probability of choosing a prime number is: P(prime) = n(prime)n(S) = (4)/(7) The probability of choosing a prime number is (4)/(7). Step 3: Find the probability of choosing a number greater than 10. The numbers in S greater than 10 are \11, 14, 17\. The number of elements greater than 10 is n(>10) = 3. The probability of choosing a number greater than 10 is: P(>10) = (n(>10))/(n(S)) = (3)/(7) The probability of choosing a number greater than 10 is (3)/(7). Step 4: Find the probability of choosing a multiple of 3. The multiples of 3 in S are \3, 6\. The number of multiples of 3 is n(multiple of 3) = 2. The probability of choosing a multiple of 3 is: P(multiple of 3) = n(multiple of 3)n(S) = (2)/(7) The probability of choosing a multiple of 3 is (2)/(7). Part 11 (b): A bag contains 4 blue balls, 3 green balls, and 3 yellow balls. The total number of balls is 4 + 3 + 3 = 10. One ball is drawn at random and not replaced. A second ball is then drawn. Step 1: Find the probability of obtaining two yellow balls. The probability of drawing a yellow ball first is (3)/(10). After drawing one yellow ball, there are 2 yellow balls left and 9 total balls left. The probability of drawing a second yellow ball is (2)/(9). The probability of drawing two yellow balls is: P(two yellow balls) = (3)/(10) × (2)/(9) = (6)/(90) = (1)/(15) The probability of obtaining two yellow balls is (1)/(15). Step 2: Find the probability of obtaining two blue balls or two green balls. The probability of drawing two blue balls is: P(two blue balls) = (4)/(10) × (3)/(9) = (12)/(90) = (2)/(15) The probability of drawing two green balls is: P(two green balls) = (3)/(10) × (2)/(9) = (6)/(90) = (1)/(15) The probability of drawing two blue balls or two green balls is the sum of their individual probabilities: P(two blue or two green balls) = P(two blue balls) + P(two green balls) P(two blue or two green balls) = (2)/(15) + (1)/(15) = (3)/(15) = (1)/(5) The probability of obtaining two blue balls or two green balls is (1)/(5). Step 3: Find the probability of obtaining one blue ball and one green ball in any order. This means either (blue first, then green) or (green first, then blue). The probability of drawing a blue ball first, then a green ball: P(B_1 and G_2) = (4)/(10) × (3)/(9) = (12)/(90) = (2)/(15) The probability of drawing a green ball first, then a blue ball: P(G_1 and B_2) = (3)/(10) × (4)/(9) = (12)/(90) = (2)/(15) The probability of obtaining one blue ball and one green ball in any order is the sum of these probabilities: P(one blue and one green) = P(B_1 and G_2) + P(G_1 and B_2) P(one blue and one green) = (2)/(15) + (2)/(15) = (4)/(15) The probability of obtaining one blue ball and one green ball in any order is (4)/(15). Send me the next one 📸