One hundred tickets are sold for a raffle. Three prizes are awarded.
What is the probability that Andy wins first prize, Brianna wins second, and Cam wins third?
What is the probability that Andy, Brianna, and Cam are the three winners (not necessarily in that order)?
Prob(Andy, Briana, Cam)
= (1/100)(1/99)(1/98) = 1/970200
prob (those 3 win)
= 1/C(100,3)
= 1/161700
Thank you!!!!!!!
To calculate the probability of specific individuals winning in a raffle, we need to consider the total number of possible outcomes and the number of favorable outcomes.
For the first question:
1. The probability of Andy winning first prize is 1 out of 100, as there is one winning ticket and 100 total tickets.
2. After Andy wins the first prize, there are 99 tickets remaining for the second prize. Therefore, the probability of Brianna winning second prize is 1 out of 99.
3. After Andy and Brianna have won the first two prizes, there are 98 tickets remaining for the third prize. Therefore, the probability of Cam winning third prize is 1 out of 98.
To find the probability of all three events occurring together, we multiply the individual probabilities:
Probability of Andy winning first prize = 1/100
Probability of Brianna winning second prize = 1/99
Probability of Cam winning third prize = 1/98
Probability of Andy, Brianna, and Cam winning in the given order = (1/100) * (1/99) * (1/98) = 1/970200
Therefore, the probability that Andy wins first prize, Brianna wins second prize, and Cam wins third prize is 1/970200.
For the second question:
To find the probability that Andy, Brianna, and Cam are the three winners (not necessarily in that order), we need to consider the different possible arrangements of the three winners.
There are 3! = 3 * 2 * 1 = 6 possible arrangements (permutations) for the three winners. They could be in any order: ABC, ACB, BAC, BCA, CAB, CBA.
The probability of any specific arrangement occurring is the same as in the previous calculation: 1/970200.
Since there are 6 possible arrangements, we multiply the probability of one specific arrangement by 6:
Probability of Andy, Brianna, and Cam winning in any order = (1/970200) * 6 = 1/161700
Therefore, the probability that Andy, Brianna, and Cam are the three winners (not necessarily in that order) is 1/161700.