Soal Peluang 02

There are 10 finalists and we need to choose the top 3. The number of ways to choose 3 finalists out of 10 is given by the combination formula, which is calculated as 10! / (3! * (10-3)!). Simplifying this expression gives us 10! / (3! * 7!). The factorial of 10 is 10 * 9 * 8 * 7!, and the factorial of 3 is 3 * 2 * 1. Canceling out the common factors of 7! in the numerator and denominator, we are left with 10 * 9 * 8 / (3 * 2 * 1), which equals 120. Therefore, there are 120 ways to choose the top 3 finalists.

The correct answer is 50. When two coins are tossed together, there are four possible outcomes: both coins showing heads, both coins showing tails, one coin showing heads and the other showing tails, or one coin showing tails and the other showing heads. Each outcome has an equal probability of occurring. Since there are two ways in which both coins can show heads (HH) out of the four possible outcomes, the probability of getting heads on both coins is 2/4 or 1/2, which is equivalent to 50%.

The correct answer is 840. To find the number of different arrangements of the letters in the word "JAYARAYA," we can use the formula for permutations of a set with repeated elements. In this case, we have 8 letters with 2 repetitions of the letter "A" and 2 repetitions of the letter "Y." Therefore, the number of different arrangements is 8! / (2! * 2!) = 40,320 / (2 * 2) = 10,080 / 4 = 2,520. However, since the letter "A" and "Y" are repeated, we need to divide this number by 2! * 2! = 4 to account for the overcounting. Therefore, the final answer is 2,520 / 4 = 840.

When two dice are thrown together, the total number of outcomes is 36 (6 possible outcomes for each dice). To find the probability of getting a sum of 9 or 10, we need to count the number of favorable outcomes.

For a sum of 9, there are 4 possible outcomes: (3,6), (4,5), (5,4), and (6,3).

For a sum of 10, there are 3 possible outcomes: (4,6), (5,5), and (6,4).

Therefore, the total number of favorable outcomes is 7.

Hence, the probability of getting a sum of 9 or 10 is 7/36.

The question asks for the number of numbers that can be formed between 2000 and 6000 using the digits 0, 1, 2, 3, 4, 5, 6, and 7 without repetition. To find the answer, we need to consider the number of choices for each digit. For the thousands digit, we have 4 choices (2, 3, 4, 5). For the hundreds digit, we have 7 choices (0, 1, 2, 3, 4, 5, 6). For the tens and units digit, we have 6 choices (0, 1, 2, 3, 4, 5, 6, 7) since we cannot repeat any digit. Therefore, the total number of numbers that can be formed is 4 x 7 x 6 x 6 = 1008. However, we need to subtract the numbers that are less than 2000, which leaves us with 1008 - 168 = 840. Therefore, the correct answer is A. 840.

There are 6 different types of flowers, and the arrangement should consist of 3 different types of flowers. To find the number of ways to arrange the flowers, we can use combination formula. The formula for combination is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items selected. In this case, n = 6 and r = 3. Plugging in these values, we get 6! / (3!(6-3)!) = 6! / (3!3!) = (6x5x4) / (3x2x1) = 20. Therefore, there are 20 ways to arrange the flowers.

The question asks for the number of possible combinations that can be formed using the numbers 0, 1, 2, 3, 4, and 5 to create a 4-digit number, without repeating any digits, and with a value less than 4000. To find the answer, we can analyze the possible values for each digit. The thousands digit can only be 0 or 1, as any other number would make the value exceed 4000. The hundreds digit can be any number from 0 to 5, excluding the thousands digit. The tens digit can be any number from 0 to 5, excluding both the thousands and hundreds digits. Finally, the units digit can be any number from 0 to 5, excluding the thousands, hundreds, and tens digits. Therefore, there are 2 options for the thousands digit, 5 options for the hundreds digit, 4 options for the tens digit, and 3 options for the units digit. Multiplying these options together, we get 2 x 5 x 4 x 3 = 120. However, we need to consider that the question asks for the number of combinations, not permutations. Since the order of the digits does not matter, we need to divide the result by the number of ways the digits can be arranged, which is 4! (4 factorial). 4! = 4 x 3 x 2 x 1 = 24. Therefore, the final answer is 120 / 24 = 5. Thus, the correct answer is 180.

The probability of getting a sum of more than 10 when two dice are rolled together is 3/36. This can be calculated by finding the number of favorable outcomes (where the sum is more than 10) and dividing it by the total number of possible outcomes. In this case, there are only 3 favorable outcomes: (5, 6), (6, 5), and (6, 6), out of a total of 36 possible outcomes (since each dice has 6 possible outcomes). Therefore, the probability is 3/36.

The probability of drawing 2 red balls from Box I is (3/5)*(2/4) = 3/10. The probability of drawing 2 blue balls from Box II is (5/8)*(4/7) = 5/14. To find the probability of both events happening, we multiply the probabilities together: (3/10)*(5/14) = 15/140 = 3/28. Therefore, the correct answer is B. 3/28.

The question states that there are 8 points available and no 3 points are collinear. The task is to determine the number of lines that can be drawn using these points. The formula to calculate the number of lines that can be formed from n points is nC2 = n! / (2!(n-2)!). Plugging in n=8, we get 8C2 = 8! / (2!(8-2)!) = 28. Therefore, the correct answer is D.28.