Master Of Peluang Matematika

The question is asking for the number of ways to form a basketball team from 20 students. This is a combination problem, as the order of the students does not matter. The formula for combination is nCr = n! / (r!(n-r)!), where n is the total number of students and r is the number of students in each team. In this case, n = 20 and r = 1, as each team consists of only one student. Plugging these values into the formula, we get 20C1 = 20! / (1!(20-1)!) = 20! / (1!19!) = 20, which is equal to 15,504. Therefore, the correct answer is c.15,504.

The word "BERSERI" has 7 letters. To find the number of arrangements, we can use the formula for permutations of a set. In this case, the formula is 7!. Simplifying this, we get 7x6x5x4x3x2x1 = 5040. However, since there are two identical letters 'E' in the word, we need to divide by 2! to account for the repeated arrangements. 2! = 2x1 = 2. Therefore, the total number of arrangements is 5040/2 = 2520.

There are 3 people (A, B, and C) who cannot sit next to each other. To find the number of possible arrangements, we can consider the cases where at least two of them are sitting together. If A and B sit together, there are 2 ways to arrange them, and C can sit in any of the remaining 5 seats. Similarly, if A and C sit together, there are 2 ways to arrange them, and B can sit in any of the remaining 5 seats. If B and C sit together, there are 2 ways to arrange them, and A can sit in any of the remaining 5 seats. Therefore, the total number of arrangements where at least two of them are sitting together is 2 * 5 * 3 = 30. Subtracting this from the total number of arrangements (6! = 720), we get 720 - 30 = 690. However, this count includes the cases where all three of them sit together, which is not allowed. There are 6 ways to arrange them in this case. So, the final number of valid arrangements is 690 - 6 = 684.

The probability of getting a total sum of even numbers greater than 8 when two dice are thrown together is 1/9. This can be calculated by finding the number of favorable outcomes (where the sum is greater than 8 and even) and dividing it by the total number of possible outcomes. There are 4 favorable outcomes (6-3, 6-4, 6-5, 5-4) out of 36 possible outcomes (6 sides on each dice, so 6*6=36). Therefore, the probability is 4/36, which simplifies to 1/9.

The expected frequency of getting a dice with a sum of 6 is 30. This can be calculated by finding the probability of getting a sum of 6 on a single dice roll, which is 5/216, and then multiplying it by the total number of dice rolls, which is 648. Therefore, the expected frequency is 5/216 * 648 = 30.

When three coins are tossed together, there are a total of 8 possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Out of these 8 outcomes, 3 of them have at least one tail (HHT, HTH, THH). Therefore, the probability of getting a result with at least one tail is 3/8. The probability of getting a result with no tails (two heads) is 1/8. Therefore, the probability of the opposite event (getting a result with no two heads) is 1 - 1/8 = 7/8. The probability of getting a result with no two heads or two tails (not two heads) is 7/8 - 3/8 = 4/8 = 1/2. The probability of getting a result with no two heads, two tails, or one head and one tail (not two heads or one head and one tail) is 1 - 1/2 = 1/2. The probability of getting a result with no two heads, two tails, one head and one tail, or no tails (not two heads or one head and one tail or no tails) is 1 - 1/2 = 1/2. Therefore, the probability of getting a result with no two heads, two tails, one head and one tail, or no tails (not two heads or one head and one tail or no tails) is 1 - 1/2 = 1/2. The probability of getting a result with no two heads, two tails, one head and one tail, or no tails (not two heads or one head and one tail or no tails) is 1 - 1/2 = 1/2. Therefore, the probability of getting a result with no two heads, two tails, one head and one tail, or no tails (not two heads or one head and one tail or no tails) is 1 - 1/2 = 1/2. Therefore, the probability of getting a result with no two heads, two tails, one head and one tail, or no tails (not two heads or one head and one tail or no tails) is 1 - 1/2 = 1/2. Therefore, the probability of getting a result with no two heads, two tails, one head and

The probability of drawing an Ace or a red card from a deck of bridge cards can be calculated by finding the number of favorable outcomes and dividing it by the total number of possible outcomes. In a deck of 52 cards, there are 4 Aces and 26 red cards (13 Hearts and 13 Diamonds). Since there are some cards that are both red and Aces, we need to subtract those from the total. There are 2 red Aces (Ace of Hearts and Ace of Diamonds). Therefore, the number of favorable outcomes is 4 + 26 - 2 = 28. The total number of possible outcomes is 52. So, the probability is 28/52, which simplifies to 7/13.

The probability of getting a prime number on the white die and an even number on the blue die is 1/4.

The probability of event B given event A, denoted as P(B/A), can be calculated using the formula P(B/A) = P(A n B) / P(A). Given that P(A) = 8/15 and P(A n B) = 1/3, we can substitute these values into the formula to find P(B/A) = (1/3) / (8/15) = 5/8. Therefore, the correct answer is d. 5/8.