Question

instrument # of students guitar 13 bass 10 drums 10 keyboard 2 instrument # of students guitar 3 bass 8 drums 8 keyboard 6 based on these results, express the probability that a seventh grader chosen at random will play an instrument other than keyboard as a fraction in simplest form. 33/35 8/60 52/60 20/33

Explanation:

Step1: Calculate total students

First, sum the number of students for each instrument across both tables. For the first table: Guitar (13) + Bass (10) + Drums (10) + Keyboard (2) = 35. For the second table: Guitar (3) + Bass (8) + Drums (8) + Keyboard (6) = 25. Total students = 35 + 25 = 60.

Step2: Calculate students not playing keyboard

Subtract keyboard players from total. Keyboard players: 2 + 6 = 8. So non - keyboard players = 60 - 8 = 52. Wait, no, wait. Wait, another way: sum non - keyboard in each table. First table non - keyboard: 13 + 10 + 10 = 33. Second table non - keyboard: 3 + 8 + 8 = 19. Total non - keyboard: 33+19 = 52? Wait, no, total students is 35 + 25 = 60. Keyboard is 2 + 6 = 8. So non - keyboard is 60 - 8 = 52? Wait, but 52/60 simplifies? Wait, no, wait the first table: Guitar 13, Bass 10, Drums 10, Keyboard 2. Second table: Guitar 3, Bass 8, Drums 8, Keyboard 6. So total Guitar: 13 + 3 = 16, Bass: 10 + 8 = 18, Drums: 10 + 8 = 18, Keyboard: 2+6 = 8. Total students: 16 + 18 + 18 + 8 = 60. Non - keyboard: 16+18+18 = 52. Wait, but 52/60 simplifies? Wait, no, the options have 33/35, 8/60, 52/60, 20/33. Wait, maybe I made a mistake. Wait, maybe the two tables are two different groups? Wait, no, the problem says "seventh grader chosen at random", so total students are sum of both tables. Wait, but let's recalculate: first table: 13 + 10 + 10 + 2 = 35. Second table: 3 + 8 + 8 + 6 = 25. Total students: 35 + 25 = 60. Keyboard players: 2 + 6 = 8. So non - keyboard: 60 - 8 = 52. So probability is 52/60? But wait, 52 and 60 have a common factor of 4. 52 ÷ 4 = 13, 60 ÷ 4 = 15. Wait, no, 52/60 simplifies to 13/15? But that's not an option. Wait, maybe the two tables are for two different groups, but the question is about seventh graders, maybe the first table is one group and the second is another? Wait, no, the problem says "based on these results", so maybe the two tables are combined. Wait, maybe I misread the tables. Let me check again. First table: Guitar 13, Bass 10, Drums 10, Keyboard 2. Second table: Guitar 3, Bass 8, Drums 8, Keyboard 6. So total students: 13 + 10 + 10 + 2+3 + 8 + 8 + 6 = 60. Keyboard: 2 + 6 = 8. Non - keyboard: 60 - 8 = 52. So 52/60. But 52/60 can be simplified? Wait, 52 and 60 are both divisible by 4: 52 ÷ 4 = 13, 60 ÷ 4 = 15. But 13/15 is not an option. Wait, the options are 33/35, 8/60, 52/60, 20/33. Wait, maybe the two tables are not combined? Maybe the first table is one group, and the second is another? Wait, no, the problem says "a seventh grader chosen at random", so total population is sum of both. Wait, maybe I made a mistake in adding. Let's add first table: 13 + 10 = 23, 23 + 10 = 33, 33 + 2 = 35. Second table: 3 + 8 = 11, 11 + 8 = 19, 19 + 6 = 25. 35 + 25 = 60. Keyboard: 2 + 6 = 8. Non - keyboard: 60 - 8 = 52. So 52/60. But 52/60 is an option (the third option). Wait, but maybe the question is about only one table? Wait, the first table: 35 students, keyboard 2. So non - keyboard: 35 - 2 = 33. Then probability 33/35. Oh! Maybe the two tables are two different questions, but the problem is about one of them? Wait, the problem says "seventh grader", maybe the first table is seventh graders and the second is eighth? The problem says "seventh grader", so maybe only the first table? Let's check: first table: Guitar 13, Bass 10, Drums 10, Keyboard 2. Total seventh graders: 13 + 10 + 10 + 2 = 35. Non - keyboard: 35 - 2 = 33. So probability 33/35. That's the first option. Ah, I see, I misread the problem. It's about seventh graders, so maybe the first table is seventh and the second is eighth. So total seventh graders: 13 + 10 + 10 + 2 = 35. Non - keyboard: 35 - 2 = 33. So probability 33/35.

Answer:

A. 33/35