Question

There are 40 students,

10 have blue eyes, 3 have green and 27 have brown.
In class 12B there are 50 students,
23 have blue eyes, 4 have green and 23 have brown.
One person is chosen at random from
each class to represent the sixth form.
What is the probability that they
both have brown eyes?

Answer 1

`(621)/(2000)` ≈ 0.31 = 31 %

set up equation

the probability of someone from the first group having brown eyes is
`(students-with-brown-eyes-in-first-group)/(total-students-in-first-group )`

the probability of someone from the second group having brown eyes is
`(students-with-brown-eyes-in-second-group)/(total-students-in-second-group )`

so the probability of both students having brown eyes is
`(students-with-brown-eyes-in-first-group)/(total-students-in-first-group )`*
`(students-with-brown-eyes-in-second-group)/(total-students-in-second-group )`

values

students with brown eyes in first group = 27

total students in first group = 40

students with brown eyes in second group = 23

total students in second group = 50

plug in values and solve

`(students-with-brown-eyes-in-first-group)/(total-students-in-first-group )`*
`(students-with-brown-eyes-in-second-group)/(total-students-in-second-group )`

=
`(27)/(40) *(23)/(50)`

=
`(27 * 23)/(40 * 50)`

=
`(621)/(2000)`

≈ 0.31

= 31 %