Question

16. In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

Answer 1

Answer:

[tex[ two groups=-68+25+25+34 -6=10[/tex]

Step-by-step explanation:

For this case we have a diagram of the situation on the figure attached.

And for this case we can us the following rule from probability:

$P(AUBUC) = P(A)+P(B) +P(C) -P(A and B)- P(A andC)-P(B and C) +P(A and B and C)$

Or equivalently:

$total= A+ B +C - two groups -2*[people in3 groups]+ nonegroups$

We know the total on this case total=68, the people for history is A=25, for math B=25 and for english C=34, and the people in all the groups 3, so we can replace:

$68=25+25+34- two groups -2*3 +0$

And if we solve for the poeple in two groups we got:

[tex[ two groups=-68+25+25+34 -6=10[/tex]

Answer 2

Answer:

Step-by-step explanation:

Total number of students = 68

Let history = H

Maths = M

English = E

n(H) = 25

n(M) = 25

n(E) = 34

n(HnMnE) = 3

Total = n(H) + n(M) + n(E) - people in exactly two groups + 2(people in exactly 3 groups) + people in none of the groups

68 = 25 + 25 + 34 - people in exactly two groups - 6 +0

68 = 84 -6 - people in exactly two groups

68 = 78 - people in exactly two groups

People in exactly two groups = 78 - 68

= 10

OR

From the venn diagram, people in exactly two groups are represented by x, y and z

Total = 25 - x - y - 3 + 25 - x - z - 3 + 34 - y - z - 3 + x + y + z + 3

68 = 50 - x - 3 + 34 - y - z - 3

68 = 84 - 6 - x - y - z

68 = 78 - x - y - z

68 - 78 = - x - y - z

-10 = -(x + y + z)

x+y+z = -10/-1

x+y+z = 10

The number of students that registered for exactly two courses = 10