Question

A high school is offering 3 classes: math, physics, and chemistry. The classes are open to any of the 100 students in the school. There are 28 students in the math class, 26 in the physics class, and 16 in the chemistry class. There are 12 students who are in both math and physics, 4 who are in both math and chemistry, and 6 who are in both physics and chemistry. In addition, there are 2 students taking all 3 classes. If a student is chosen randomly, what is the probability that he or she is not in any of the classes?

Answer 1

Answer: Student is chosen randomly than probability of every student is same to chosen = $\frac{50}{100}$

Step-by-step explanation:

No of student = 100

Let A denote Math class , B denote Physics class , C denote Chemistry class

n(A) = 28 , n(B) = 26 , n( C)= 16

Student both in math and physics class = $n(A\bigcup B) = 12$

Student both in math and chemistry = $n(A\bigcup C) = 4$

Student both in physics and chemistry = $n(B\bigcup C) = 6$

Student taking all classes = $n(A\bigcap B\bigcap c) = 2$

Student is chosen randomly than probability of every student is same to chosen

probability of 1 student being chosen = $\frac{1}{100}$

student is chosen randomly then probability of student he or she not taking any classes

but from the Venn diagram the probability is = $\frac{50}{100}$

$P(A\bigcup B\bigcup C)^{c} = 1-P((A\bigcup B\bigcup C))$

$P((A\bigcup B\bigcup C)) = P(A) + P(B) + P(C) - P(A\bigcap B) - P(B\bigcap C) - P(A\bigcap C) + P(A\bigcap B\bigcap c) ( using venn diagram)$

$= \frac{28}{100} + \frac{26}{100} + \frac{16}{100} - \frac{12}{100} - \frac{4}{100} - \frac{6}{100} + \frac{2}{100}$

$= \frac{50}{100}$

$P(A\bigcup B\bigcup C)^{c} = 1-P((A\bigcup B\bigcup C))$

$= 1 - \frac{50}{100} = \frac{50}{100}$