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
Step 1: Analyze the given set for question 11(a). The set is S = \3, 5, 6, 8, 11, 14, 17\. The total number of elements in the set is n(S) = 7. (i) Probability that the number is even. The even numbers in S are \6, 8, 14\. The number of even numbers is n(even) = 3. The probability is: P(even) = n(even)n(S) = (3)/(7) The probability that the number is even is (3)/(7). (ii) Probability that the number is prime. The prime numbers in S are numbers greater than 1 with only two divisors: 1 and themselves. The prime numbers in S are \3, 5, 11, 17\. The number of prime numbers is n(prime) = 4. The probability is: P(prime) = n(prime)n(S) = (4)/(7) The probability that the number is prime is (4)/(7). (iii) Probability that the number is 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 is: P(>10) = (n(>10))/(n(S)) = (3)/(7) The probability that the number is greater than 10 is (3)/(7). (iv) Probability that the number is 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 is: P(multiple of 3) = n(multiple of 3)n(S) = (2)/(7) The probability that the number is a multiple of 3 is (2)/(7). Step 2: Analyze the bag contents for question 11(b). A bag contains: • 4 blue balls (B) • 3 green balls (G) • 3 yellow balls (Y) Total number of balls = 4 + 3 + 3 = 10. One ball is drawn and not replaced, then a second ball is drawn. (i) Probability of obtaining two yellow balls. Probability of drawing a yellow ball first (Y_1): P(Y_1) = Number of yellow ballsTotal balls = (3)/(10) After drawing one yellow ball, there are 2 yellow balls left and 9 total balls. Probability of drawing a second yellow ball (Y_2) given the first was yellow: P(Y_2 | Y_1) = Number of yellow balls remainingTotal balls remaining = (2)/(9) Probability of two yellow balls: P(Y_1 and Y_2) = P(Y_1) × P(Y_2 | Y_1) = (3)/(10) × (2)/(9) = (6)/(90) = (1)/(15) The probability of obtaining two yellow balls is (1)/(15). (ii) Probability of obtaining two blue balls or two green balls. This is P(two blue balls) + P(two green balls). Probability of two blue balls (B_1 and B_2): P(B_1) = (4)/(10) P(B_2 | B_1) = (3)/(9) P(B_1 and B_2) = (4)/(10) × (3)/(9) = (12)/(90) = (2)/(15) Probability of two green balls (G_1 and G_2): P(G_1) = (3)/(10) P(G_2 | G_1) = (2)/(9) P(G_1 and G_2) = (3)/(10) × (2)/(9) = (6)/(90) = (1)/(15) Probability of two blue balls or two green balls: P(two blue or two green) = P(B_1 and B_2) + P(G_1 and G_2) = (2)/(15) + (1)/(15) = (3)/(15) = (1)/(5) The probability of obtaining two blue balls or two green balls is (1)/(5). (iii) Probability of obtaining one blue ball and one green ball in any order. This means P(B_1 and G_2) + P(G_1 and B_2). Probability of drawing a blue ball first, then a green ball (B_1 and G_2): P(B_1) = (4)/(10) After drawing one blue ball, there are 3 green balls left and 9 total balls. P(G_2 | B_1) = (3)/(9) P(B_1 and G_2) = (4)/(10) × (3)/(9) = (12)/(90) = (2)/(15) Probability of drawing a green ball first, then a blue ball (G_1 and B_2): P(G_1) = (3)/(10) After drawing one green ball, there are 4 blue balls left and 9 total balls. P(B_2 | G_1) = (4)/(9) P(G_1 and B_2) = (3)/(10) × (4)/(9) = (12)/(90) = (2)/(15) Probability of one blue ball and one green ball in any order: P(one blue and one green) = P(B_1 and G_2) + P(G_1 and B_2) = (2)/(15) + (2)/(15) = (4)/(15) The probability of obtaining one blue ball and one green ball in any order is (4)/(15). 3 done, 2 left today. You're making progress.