A. 4 cara D. 18 cara B. 6 cara E. 24 cara C.10 cara 33. Dari 10 finalis lomba AFI akan dipilih juara I,II, dan III. Banyaknya kemungkinan susunan terpilihnya sebagai juara adalah __ A C. 480 E 720 B. 240 D. 620 34. Sebuah kompetisi sepak bola Eropa "EURO"diikuti oleh 6 negara. Pada babak awal setiap negara harus bertanding satu sama lain Banyaknya pertanding- an pada babak awal adalah __ A. 36 C. 15 E 6 B. 30 D. 12 35. Sebuah kotak berisi 5 kelereng merah dan 3 kelereng kuning. Jika diambil dua kelereng secara acak satu per satu berturut-turut tanpa pengembalian, maka peluang terambil pertama kelereng merah dan kedua kelereng kuning adalah __ A. (3)/(4) (5)/(14) E (15)/(64) B. (8)/(15) D. (15)/(56)

Let's solve each question one by one:<br /><br />**33. Selecting winners from 10 finalists:**<br /><br />To determine the number of possible arrangements for selecting a winner for positions I, II, and III from 10 finalists, we use permutations because the order matters.<br /><br />The formula for permutations is given by:<br />\[ P(n, r) = \frac{n!}{(n-r)!} \]<br /><br />Where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.<br /><br />Here, \( n = 10 \) and \( r = 3 \).<br /><br />\[ P(10, 3) = \frac{10!}{(10-3)!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720 \]<br /><br />So, the number of possible arrangements is **720**.<br /><br />**Answer: E. 720**<br /><br />---<br /><br />**34. Number of matches in a round-robin tournament with 6 teams:**<br /><br />In a round-robin tournament, each team plays every other team exactly once. The number of matches can be calculated using combinations since the order of matches does not matter.<br /><br />The formula for combinations is:<br />\[ C(n, 2) = \frac{n(n-1)}{2} \]<br /><br />Where \( n \) is the number of teams.<br /><br />For 6 teams:<br />\[ C(6, 2) = \frac{6 \times 5}{2} = 15 \]<br /><br />So, there are **15** matches in the initial round.<br /><br />**Answer: C. 15**<br /><br />---<br /><br />**35. Probability of drawing a red marble first and a yellow marble second without replacement:**<br /><br />First, calculate the probability of drawing a red marble first. There are 5 red marbles out of a total of 8 marbles.<br /><br />Probability of first red marble:<br />\[ P(\text{Red first}) = \frac{5}{8} \]<br /><br />After drawing a red marble, there are now 4 red marbles and 3 yellow marbles left, making a total of 7 marbles.<br /><br />Probability of second yellow marble:<br />\[ P(\text{Yellow second | Red first}) = \frac{3}{7} \]<br /><br />The combined probability of both events happening in sequence is:<br />\[ P(\text{Red first and Yellow second}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56} \]<br /><br />So, the probability is \(\frac{15}{56}\).<br /><br />**Answer: D. \(\frac{15}{56}\)**