Download Sample Teacher`s Solution

Sample Teacher’s Solution
Problem
A 6-sided die is rolled at the same time that a coin is tossed.
1. Write a sample space in which the outcomes are equally likely. Explain
why they are equally likely.
2. Make a tree diagram or an area model for the situation.
3. Find the probability of getting an outcome in which the die’s number is
divisible by 3 and the coin is heads. Explain or illustrate how you can get
this answer using the sample space and your model.
Solution
1. We start by considering the sample spaces of the die roll and coin toss
separately. The finest sample space for a die roll is the sample space
{1, 2, 3, 4, 5, 6}; the only nontrivial sample space for a coin toss is {H, T }.
The die and coin are both assumed to be fair, so all outcomes in these
sample spaces are equally likely.
Because the die roll and coin toss are independent, any outcome of the
die roll can be paired with any outcome of the coin toss. This leads us to
write the following sample space:
(1, H) (2, H) (3, H) (4, H) (5, H) (6, H)
Ω=
(1, T ) (2, T ) (3, T ) (4, T ) (5, T ) (6, T )
Every possible result of the die roll and coin flip is encoded in one of the
outcomes. On the other hand, no possible result of the die roll and coin
flip is encoded by more than one outcome: if so, either the die would have
to take two numbers or the coin would have to be both heads and tails.
Because of these two facts, we know Ω is a valid sample space.
Furthermore, all outcomes in Ω are equally likely because the die roll and
coin toss are independent. More precisely, the probability of (1, H) and
(1, T ) are the same because H and T are equally likely outcomes of the
coin toss, independently of the fact that the die reads 1. Similarly, (1, H)
and (2, H) are equally likely because 1 and 2 are equally likely outcomes of
the die roll, independently of the fact that the coin landed on heads. This
1
reasoning can be extended to show the likelihood of each of the outcomes
is the same.
2. Conceptually, we can think of the process of rolling the die and tossing
the coin as separate trails. Thinking of the coin as being flipped before
the die is rolled, we get the following tree diagram:
·
1
2
1
2
H
1
6
(1,H)
(2,H)
1
6
(3,H)
1
6
T
1
6
1
6
(4,H)
1
6
1
6
(5,H)
(6,H)
(1,T )
(2,T )
1
6
1
6
(3,T )
1
6
1
6
(4,T )
(For those who are curious, it is not wrong to use an area mode here. However, there are some times when one is better than the other.)
3. We examine Ω and identify all outcomes for which the die’s number is
divisible by 3 and the coin is heads. They are: {(3, H), (6, H)}. The
probability of the event that the head’s number is divisible by 3 and the
coin is heads is equal to the sum of the probabilities of the outcomes
for which this is true. The probability of each outcome is given by the
product of the probabilities of taking each of the branches in the tree
diagram leading up to it. We have:
P (die’s number divisible by 3 and H) = P ((3, H))+P ((6, H)) =
2
1 1 1 1
1
· + · =
2 6 2 6
6
1
6
(5,T )
(6,T )