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There are 9 people taking part in a raffle.
Ann, Bob, Hans, Jim, Kira, Lena, Omar, Ravi, and Soo.
Suppose that prize winners are randomly selected from the 9 people.
Compute the probability of each of the following events.
Event A: Bob is the first prize winner, Lena is second, and Ann is third,
Event B: The first three prize winners are Soo, Omar, and Kira, regardless of order
Write your answers as fractions in simplest form.
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Answer:

  • Probability of event A: 1/504
  • Probablity of event B: 1/84

Explanation:

1. Event A: Bob is the first prize winner, Lena is second, and Ann is third,

  • The probability that Bob is the first prize winner is one outcome out of 9 possible outcomes, so it is 1/9

  • The probability that Lena is second, after Bob is the first prize winner is one outocome out of 8 possible outocmes, so it is 1/8

  • The probability Ann is third, after Bob y first and Lena is second is 1/7.

Thus, the joint probability that Bob is the first prize winner, Lena is second, and Ann is third is the product of the three calculated probabilities:

  • 1/9 × 1/8 × 1/7 = 1 / (9×8×7) = 1/504 ← answer

2. Event B: The first three prize winners are Soo, Omar, and Kira, regardless of order

When the order does not matter, the number of combinations for three different persons win the 3 prizes is C(3,3), which is computed with the corresponding formula:

          [tex]C(m,n)=\dfrac{m!}{n!(m-n)!}[/tex]  

Thus:

          [tex]C(3,3)=\dfrac{3!}{(3!(3-3)!}=1[/tex]

And the number of possible combinations of winners is C(9,3):

                 [tex]C(9,3)=\dfrac{9!}{3!(9-3)!}=84[/tex]

Then, the probability is the number of favorable combinations, C(3,3) = 1, divided by the number of possible combinations, C(9,3) = 84:

  • 1/84 ← answer

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