# Dice Worksheet by ajy78640

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```									           Basic Probability Worksheet Answers – Introduction to Statistics

1. Roll 2 fair 4-sided dice. What is the probability that the sum of the dice is 2? 3? 4? 5? 6?
7? 8? (Hint: Tree diagram may be helpful)

The sample space is {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4),
(4, 1), (4, 2), (4, 3), (4, 4)}. Using this we have:

sum             prob
2              1/16
3              2/16
4              3/16
5              4/16
6              3/16
7              2/16
8              1/16

2. From a standard deck of cards, you are dealt 5 cards. What is the probability that:
13       12       11       10           9
a) all of your cards are hearts?:      52
·   51
·   50
·   49
·   48
≈ .000495. (Multiplication Rule)
39       38       37       36       35
b) at least one of your cards is a heart?: 1 − Prob(no heart) = 1 −                                                 52
·   51
·   50
·   49
·   48
≈ .7785
(Not Rule and Multiplication Rule)
52          48       44       40       36
c) all of your cards have diﬀerent values?:                   52
·   51
·   50
·   49
·   48
≈ .507. (Multiplication Rule)

3. Roll 2 fair 6-sided dice, one at a time. What is the probability that:

a) you get doubles (both dice show the same value)?: 36 possible choices from our sample space
6
(using the tree diagram approach), 6 of which are doubles. the answer is 36 .
1
b) the dice show diﬀerent values?: 1 − Prob(doubles) = 1 −                                                6
= 5 . (Not Rule)
6
3        3
c) the ﬁrst is odd and the second is even?                    6
·    6
= 1 . (Multiplication rule)
4

d) the ﬁrst is odd or the second is even?: Let X be the event that the ﬁrst is odd and Y be the
event that the second is even. We want Prob(X or Y = Prob(X)+Prob(Y )−Prob(X and Y ) =
3
6
+ 3 − 3 · 3 = 3 . (Addition Rule)
6   6 6      4

e) the ﬁrst is divisible by 3 or the sum of the dice is less than 7? Let X be the event that the
ﬁrst is divisible by 3 and Y the event that the sum of the dice is less than 7 (i.e., equals 2,3,4,5,
or 6). We want Prob(X or Y = Prob(X) + Prob(Y ) − Prob(X and Y ) = 2 + 15 − 36 = 23 . The
6  36
4
36
15                                                                                              4
36
comes from getting a sum of 2 or 3 or 4 or 5 or 6 and suing the addition rule. The 36 comes
from the rolls (3, 1), (3, 2), (3, 3), (6, 1) being the only rolls that satisfy both conditions.
4. You have 12 pairs of socks: 5 identical white pairs, 4 identical black pairs, 2 identical blue
pairs, and 1 mismatched pair. You reach in and pick 2 socks. What is the probability that:

a) you get a non-mismatched pair? You could pick two whites, two blacks, or two blues. These
are mutually exclusive events, so the probability that one of these occurs is found with the
addition rule, which reduces to the sum of the individual probabilities when we have mutually
9    8    7    4    3
exclusive events. The answer is 10 · 23 + 24 · 23 + 24 · 23 ≈ .286.
24

b) you get a pair if the ﬁrst sock you pick is white? In this case the second sock must be one of
9
the 9 remaining socks. The answer is 23 .

Now, your little brother comes in and takes one of the socks and puts it under the bed. You did
not see which sock it was. Now what is the probability that

c) you get a non-mismatched pair? It is still .286 since you really don’t have any more informa-
tion.

d) the sock under the bed is blue if the two socks you picked were both black?

Let X be the event that the sock under the bed is blue and Y the event that both socks picked
were black. We want Prob(X|Y ). Using the formula Prob(X and Y )/Prob(Y ) we ﬁnd that
4   8   7     8   7     4    2
Prob(X|Y ) = 24 · 23 · 22 / 24 · 23 = 22 = 11 . This should make sense if you argue as follows.
If you know you picked two black socks, then the sock under the bed could be one of 22 possible
socks. There are 4 blue socks, hence the 4/22 answer.

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