), so with probability 1
/
4
k
+1
the numerator and denominator before reduction were
1
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both multiples of 2
k
but nor 2
k
+1
. Summing over
k
, the probability that both the randomly
chosen numerator and denominator are odd is
1
4
+
1
4
2
+
1
4
3
+
· · ·
=
1
3
.
4.
Dungeons and Dragons players use dice in the shape of each of the regular solids. The
faces are always numbered 1 through the number of total faces there are. You shake all five
dice. what is the probability of your throwing a total of 6? (B+S 7.2.31)
Solution.
There are five outcomes that have a total of 6: we must get 2 on one die and
1 on the other four dice.
There are 4
·
6
·
8
·
12
·
20 = 46080 possible outcomes.
So the
probability of throwing a 6 is 5
/
46080 = 1
/
9216.
5.
Suppose you flip a fair coin 10 times on two different ocassions. One time you see
10 heads, the other time you see
HHTHHHTTHT
. Is either one of these outcomes more
likely than the other? Which one is random? Explain. (B+S 7.3.31)
Solution.
Neither outcome is more likely than the other; they both have probability 1
/
2
10
of occuring. Both are random! You might think the second one is “more random” then the
first.
6.
Suppose you deal three cards from a regular deck of 52 cards. What is the probability
that they will all be jacks? (B+S 7.4.11)
Solution.
The probability that the first card dealt is a jack is 4
/
52; the probability that
the second card dealt is a jack, given that the first one was a jack, is 3
/
51 (since there are
51 cards left, of which 3 are jacks); the probability that the third card dealt is a jack, given
that the first two are jacks, is 2
/
50. So the answer is
4
·
3
·
2
52
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 Summer '09
 Lugo
 Math, Probability, Elementary arithmetic, Randomness, Dice, Dungeons & Dragons

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