There are 10 people taking part in a raffle.Ann, Bob, Elsa, Hans, Jim, Kira, Lena, Omar, Ravi, and Soo.Suppose that prize winners are randomly selected from the 10 people.Compute the probability of each of the following events.Event A: Ravi is the first prize winner, Kira is second, and Lena is third.Event B: The first three prize winners are Ann, Elsa, and Lena, regardless of order.Write your answers as fractions in simplest form.

Event A: The probability of Ravi being the first prize winner, Kira being the second, and Lena being the third is calculated by multiplying the individual probabilities of each event.

The probability of Ravi being the first prize winner is 1/10, because there are 10 people in total.

After Ravi has won, there are 9 people left, so the probability of Kira being the second prize winner is 1/9.

After Ravi and Kira have won, there are 8 people left, so the probability of Lena being the third prize winner is 1/8.

So, the probability of event A is (1/10) * (1/9) * (1/8) = 1/720.

Event B: The first three prize winners are Ann, Elsa, and Lena, regardless of order. This means we have to calculate the probability of these three people winning the first three prizes in any order.

The probability of any one person winning the first prize is 1/10, the second prize is 1/9, and the third prize is 1/8, as before. So the probability of Ann, Elsa, and Lena winning the first three prizes in a specific order is (1/10) * (1/9) * (1/8) = 1/720.

However, since the order does not matter, we have to consider all the possible orders in which these three people could win. There are 3! = 321 = 6 different orders.

So, the probability of event B is 6 * (1/720) = 1/120.