Question

Can someone solve this problem for me, I really need the answer. Please make sure that it is 100% correct

At a math competition hosted at X, there are 75 students. Among these students 32 belong to X, 13 belong to Y, 19 belong to Z, and 11 belong to D. During the award ceremony, the organizers decided to randomly select a group of 10 students to form a special panel. If the panel consists of exactly 2 students from each school, what is the probability that the selected panel includes at least one student who scored in the top 50% of their respective school’s results.

Answer 1

To solve this problem, let's first calculate the total number of possible panels that can be formed from the 75 students. Since we need exactly 2 students from each school, we can calculate it as follows:

For school X: C(32, 2) - the number of ways to choose 2 students from 32.

For school Y: C(13, 2) - the number of ways to choose 2 students from 13.

For school Z: C(19, 2) - the number of ways to choose 2 students from 19.

For school D: C(11, 2) - the number of ways to choose 2 students from 11.

Let's calculate these values:

C(32, 2) = 32! / (2! * (32-2)!) = 32! / (2! * 30!) = (32 * 31) / (2 * 1) = 496

C(13, 2) = 13! / (2! * (13-2)!) = 13! / (2! * 11!) = (13 * 12) / (2 * 1) = 78

C(19, 2) = 19! / (2! * (19-2)!) = 19! / (2! * 17!) = (19 * 18) / (2 * 1) = 171

C(11, 2) = 11! / (2! * (11-2)!) = 11! / (2! * 9!) = (11 * 10) / (2 * 1) = 55

Now, let's calculate the total number of possible panels:

Total panels = C(75, 10) - the number of ways to choose any 10 students from 75.

C(75, 10) = 75! / (10! * (75-10)!) = 75! / (10! * 65!) ≈ 6.12 x 10^10

To calculate the probability of selecting at least one student who scored in the top 50% of their school, we need to calculate the total number of panels that fulfill this condition.

For school X, there are 16 students who scored in the top 50%, so we need to choose 2 students from these 16.

C(16, 2) = 16! / (2! * (16-2)!) = 16! / (2! * 14!) = (16 * 15) / (2 * 1) = 120

Similarly, for schools Y, Z, and D, the number of ways to choose 2 students from the top 50% of each school are as follows:

C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15

C(10, 2) = 10! / (2! * (10-2)!) = 10! / (2! * 8!) = (10 * 9) / (2 * 1) = 45

C(5, 2) = 5! / (2! * (5-2)!) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10

Now, let's calculate the total number of panels that include at least one student who scored in the top 50% of their school:

Total panels with at least one top 50% student = (C(16, 2) * C(6, 2) * C(10, 2) * C(5, 2)) = 120 * 15 * 45 * 10 = 972,000

Finally, let's calculate the probability:

Probability = Total panels with at least one top 50% student / Total panels = 972,000 / (6.12 x 10^10) ≈ 0.0159

Therefore, the probability that the selected panel includes at least one student who scored in the top 50% of their respective school's results is approximately 0.0159 or 1.59%.