1

I'm unsure where I'm going wrong with this.

In a class of 40 people studying music: 2 play violin, piano and recorder, 7 play at least violina nd piano, 6 play at least piano and recorder, 5 play at least recorder and violin, 17 play at least violin, 19 play at least piano, and 14 play at least recorder. How many play none of these instruments?

This is what I did:
$|A| = $ plays violin
$|B| = $ plays piano
$|C| = $ plays recorder.
From the information:
$|A\cap B\cap C| = 2 \\ |A\cup B| = 7 \\ |A\cup C| =5 \\ |B \cup C| = 6 \\ |A| = 17 \\ |B| = 19 \\ |C| = 14.$

I want $|A^c \cap B^c \cap C^c|$.
By considering
$$|A^c \cup B^c \cup C^c| = |A^c| + |B^c| + |C^c| - |A^c\cap B^c| - |A^c \cap C^c| - |B^c \cap C^c| + |A^c \cap B^c \cap C^c|\\ = (40-17)+(40-19) + (40-14) - (40-7) - (40-5) + (40-6) + |A^c \cap B^c \cap C^c|\\ = -32 + |A^c \cap B^c \cap C^c| = 40 - |A \cap B \cap C| = 38 \\ \implies |A^c \cap B^c \cap C^c| = 70...?$$

OneGapLater
  • 820
  • 1
  • 9
  • 22

2 Answers2

1

Your computations for the pairs is wrong. $|A^c\cap B^c|=40-|A\cup B|$, and you don't know $|A\cup B|$. You are computing it as $|A^c\cap B^c|=40-|A\cap B|$, which is false.

Instead, just computes $|A\cup B\cup C|$ with inclusion-exclusion, and then subtract from $40$.

Thomas Andrews
  • 177,126
0

As previously stated, take the information you know. Your events for more than 1 instrument should be intersections, not unions. That is $\cap$ not $\cup$

$|A\cap B\cap C| = 2 \\ |A\cap B| = 7 \\ |A\cap C| =5 \\ |B \cap C| = 6 \\ |A| = 17 \\ |B| = 19 \\ |C| = 14.$

You want $|A^c \cap B^c \cap C^c|$ which is simply $40-|A \cup B \cup C|$

Then for $|A \cup B \cup C|$ we have $|A| + |B| + |C| - |A \cap B| - |A\cap C| - |B \cap C| + |A \cup B \cup C|$

$=17+19+14-7-5-6+2=34$

Don't forget final step of $40-34=6$

user345
  • 771