Suppose the NBA has decided to add two more teams to the league. If 18 cities have applied for franchises, what is the probability that in a random drawing, the league selects the two westernmost cities?
Having trouble with this one.
HO HO HO, a random drawing involved in $$$ choices.
Pr=1/18*1/17
LOL..not very realistic example by the textbook.
Would you mind helping understand how you came up with that? which one of the probability formulas was used..I understand the answer just not sure how you came about it.
Thanks
To solve this problem, we need to determine the number of favorable outcomes (selecting the two westernmost cities) and the total number of possible outcomes (selecting any two cities from the 18 applicants). Then we can use probability formula to find the answer.
First, let's determine the number of favorable outcomes. Since we want to select the two westernmost cities, it means we need to find the number of ways we can select both the westernmost and second westernmost cities from the 18 applicants.
To find this, we need to understand how many cities are in the westernmost half of the group. If we arrange the cities in order from west to east, the two westernmost cities will be the first two cities in this ordered list.
Since we are selecting the first two cities, this is equivalent to counting the number of ways we can choose two cities from the westernmost half of the group. If we assume that half the cities applied from the west, then the total number of westernmost cities would be 18/2 = 9. So, we need to find how many ways we can choose two cities from this group of 9 cities.
The number of ways to choose two cities from a group of 9 can be calculated using the combination formula, which is nCr = n! / (r!(n-r)!), where n is the total number of cities and r is the number of cities we want to choose. So in this case, we have n = 9 and r = 2.
Using the combination formula, we can calculate:
9C2 = 9! / (2!(9-2)!) = 9! / (2!7!) = (9 * 8) / (2 * 1) = 36.
Therefore, there are 36 favorable outcomes (ways to select the two westernmost cities) out of all the possible combinations of selecting any two cities from the 18 applicants.
Now, let's determine the total number of possible outcomes. Since we are selecting any two cities from the 18 applicants, this can be calculated using the combination formula again, with n = 18 and r = 2:
18C2 = 18! / (2!(18-2)!) = 18! / (2!16!) = (18 * 17) / (2 * 1) = 153.
Therefore, there are 153 total possible outcomes (ways to select any two cities from the 18 applicants).
Finally, to find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes = 36 / 153 = 0.235 or approximately 23.5%.
So, the probability that in a random drawing, the NBA selects the two westernmost cities is approximately 23.5%.