A family has 3 children, 2 girls and a boy. Every morning (Tuesday through Saturday) each
child selects a stock at random from the list printed in the newspaper. If the stock is up,
that child does the morning dishes. If the stock is down, the child who picked the stock
does the noon dishes. If the stock is unchanged, that child does the evening dishes. If there
is a tie–2 children to do the noon dishes, for example–then both children work together to
do the noon dishes, and the remaining meal‘s dishes are done by default by mom or dad.
The assignments can be thought of as an ordered triple, ordered by age. For example, (+, 0,
–) means that the youngest does the morning dishes (the stock was up), the oldest child
does the noon dishes (the stock was down), and the middle child, whose stock was
unchanged, does the dishes after the evening meal. (–, 0, 0) means the youngest does the
noon dishes, the other two work together on the dishes from the evening meal. The
morning dishes are done by mom and dad. Determine the following events:
(a) E1= {the oldest and youngest work together doing the dishes; the middle child does
the dishes from a different meal}
(b) E2= {the youngest child has the day off}
(c) E3= {mom and dad do the dishes twice}
(d) E4= {the oldest child does the noon dishes alone}
(e) E‘2
(f) E3E4
(g) Let E5= {either the oldest or the youngest child does the morning dishes alone}
Let E6= {the middle child does the evening dishes alone}
E5E6
(h) E3E4
(i) E2E‘2
(j) E2E‘2
In a certain class, 40% of students had a B average at midterm. Of these, 50% ended up
with a course grade of B. Of those who did not have a B average at midterm, 40% ended up
with a course grade of B. If one of the students in the class is selected at random and is
found to have received a B for the course, what is the probability that the student did not
have a B average at midterm?