Download AMS 80A: Gambling and Gaming (Spring 2014)

AMS 80A: Gambling and Gaming (Spring 2014)
Homework 2 solutions
1. Consider placing on a bookshelf 10 books, which include 4 mathematics books, 3 chemistry books, 2
history books, and 1 language book. What is the total number of arrangements such that all books of
the same subject are placed next to each other?
Solution: In class, we have applied the multiplication counting rule to find the number of arrangements,
(4 × 3 × 2 × 1) × (3 × 2 × 1) × (2 × 1) × 1 = 288, when the math books are the first 4 on the shelf, followed
by the 3 chemistry books, the 2 history books, and the language book. Since there are (4 × 3 × 2 × 1)
orderings of the four subjects, another application of the multiplication rule yields 24 × 288 = 6912 for
the total number of arrangements.
2. Suppose that six balanced dice are rolled, and assume that all possible outcomes are equally likely. What
is the probability that each of the six numbers appears exactly once?
Solution: Applying the multiplication counting rule for both numerator and denominator, the probability can be written as (6 × 5 × 4 × 3 × 2 × 1)/66 = 0.01543.
3. Exercise 4 from “Exercise Set B”, Chapter 13 of the Freedman, Pisani and Purves book (page 227).
Solution: (a) The probability that the 5th card is the queen of spades is 1/52, since no information
is provided on the other four cards; therefore this is the same with asking for the probability that a
randomly selected card from a deck of 52 cards is the queen of spades.
(b) The probability that the 5th card is the queen of spades given that the first 4 cards are hearts is a
conditional probability equal to 1/48.
4. Five cards will be dealt off the top of a well-shuffled deck of 52 cards. Find the probability that the first
four cards are aces and the fifth card is a king? Is this the same with the probability that the first card
is a king and the other four cards are aces?
Solution: Both probabilities can be obtained using the multiplication rule for the probability of an
intersection (here, the intersection of 5 events). The probability that the first four cards are aces and the
3
2
1
4
4
× 51
× 50
× 49
× 48
, and the probability that the first card is a king and
fifth card is a king is given by 52
4
3
2
1
4
. Although the individual conditional probabilities
the other four cards are aces, 52 × 51 × 50 × 49 × 48
in the two products are not the same, the resulting probabilities for the two intersections are the same.
5. You are thinking about playing a lottery with the following rules: you buy a ticket, choose 3 different
numbers from 1 to 100, and write them on the ticket. The lottery has a box with 100 balls numbered
from 1 through 100. Three balls are drawn at random from the box without replacement. If the numbers
on the three selected balls are the same as the numbers on your ticket, you win (the order of the numbers
does not matter). If you decide to play, what is the probability of winning?
Solution: Using the multiplication rule, the probability of winning can be expressed as the product of
three probabilities, one unconditional for the first selected ball, and two conditional probabilities for the
second and third selected balls. Specifically, let p1 = Pr(first selected ball is one of the 3 on the ticket)
= 3/100, p2,1 = Pr(second selected ball is one of the two remaining on the ticket given that the first
selected ball is one of the 3 on the ticket) = 2/99, and p3,12 = Pr(third selected ball is the last remaining
on the ticket given that the first two selected balls are on the ticket) = 1/98. Therefore, Pr(winning the
lottery) = p1 × p2,1 × p3,12 = 3/100 × 2/99 × 1/98 ≈ 0.0000062.
6. Suppose that two dice are rolled repeatedly and the sum of the two numbers is recorded from each roll.
Assume that the dice are balanced, that is, all outcomes for the pair of numbers in each roll are equally
likely. Find the probability that a sum of 4 is obtained before obtaining a sum of 7.
Solution: The key idea is to recognize that this probability can be obtained as the conditional probability
of observing a sum of 4 given that the only possible outcomes are a sum of 4 or a sum of 7:
Pr(sum = 4 | sum = 4 ∪ sum = 7) =
Pr(sum = 4)
3/36
1
=
= .
Pr(sum = 4 ∪ sum = 7)
(3/36) + (6/36)
3
7. Suppose that a balanced die is rolled once, and consider the events A = (outcome is an even number), B1 =
(outcome is a number ≤ 4), and B2 = (outcome is a number ≤ 3). Are events A and B1 independent?
How about events A and B2 ?
Solution: Events A and B1 are independent, whereas events A and B2 are not independent. Both
results can be verified through the definition comparing conditional probabilities with the corresponding
unconditional probabilities; refer to the details given in class.
8. Suppose that two balanced dice are rolled independently three times. Find the probability that on each
of the three rolls, the sum of the two numbers is 7.
Solution: The probability that on each of the three rolls, the sum of the two numbers is 7 can be
expressed as Pr((7 on first roll) and (7 on second roll) and (7 on third roll)) = (Pr(sum of a roll = 7))3
= (6/36)3 , using the multiplication rule for independent events.
9. Consider three independent events A, B and C with Pr(A) = 1/4, Pr(B) = 1/3, and Pr(C) = 1/2. Obtain
the probability that none of the three events will occur. Compute also the probability that exactly one
of the three events will occur.
Solution: The probability that none of the three events will occur is given by Pr(Ac ∩ B c ∩ C c ) =
Pr(Ac ) × Pr(B c ) × Pr(C c ) = (1 − Pr(A)) × (1 − Pr(B)) × (1 − Pr(C)) = (3/4) × (2/3) × (1/2) = 1/4, where
the multiplication rule for independent events has been used to write the probability of the intersection
as the product of the corresponding probabilities.
The probability that exactly one of the three events will occur can be written as
Pr((A ∩ B c ∩ C c ) ∪ (Ac ∩ B ∩ C c ) ∪ (Ac ∩ B c ∩ C),
that is, the probability that only A will occur or only B or only C. This is a disjoint union of three
events, and therefore it can be simplified to Pr(A ∩ B c ∩ C c ) + Pr(Ac ∩ B ∩ C c ) + Pr(Ac ∩ B c ∩ C) =
((1/4) × (2/3) × (1/2)) + ((3/4) × (1/3) × (1/2)) + ((3/4) × (2/3) × (1/2)) = 11/24, using again the
multiplication rule for independent events for each of the probabilities.
10. Exercise 6 from “Exercise Set D”, Chapter 14 of the Freedman, Pisani and Purves book (pages 250,
251).
Solution: We are asked to compute Pr(A), where A = (at least one 17 in 22 throws). Again, the key idea
is to work with the complement, Ac = (no 17 in 22 throws), and use the multiplication rule for independent
events (assuming independence for the 22 throws). In particular, Pr(Ac ) = (31/32)22 = 0.497, and
therefore, Pr(A) = 0.503.
11. Every week you buy a ticket in a lottery that offers one chance out of 100,000 of winning. Assume
that lottery outcomes are independent from week to week. Obtain the probability that you will win the
lottery at least once if you play every week for five years (5 years = 260 weeks).
Solution: Pr(winning at least once in 260 weeks) = 1− Pr(never winning in 260 weeks) = 1−(99, 999/100, 000)260 ≈
0.0026, using the multiplication rule for independent events.
12. A box contains 20 red balls, 30 white balls, and 50 blue balls. Suppose that 10 balls are selected at
random one at a time, with replacement, that is, each selected ball is replaced in the box before the next
selection is made. Obtain the probability that at least one color will be missing from the 10 selected
balls.
Solution: The event of interest, A = (at least one color missing from the 10 selected balls), can be
expressed as A = A1 ∪ A2 ∪ A3 , where A1 = (no red balls selected), A2 = (no white balls selected),
and A3 = (no blue balls selected). Note that the intersection A1 ∩ A2 ∩ A3 contains no outcomes, and
therefore Pr(A1 ∪ A2 ∪ A3 ) = Pr(A1 ) + Pr(A2 ) + Pr(A3 ) − Pr(A1 ∩ A2 ) − Pr(A1 ∩ A3 ) − Pr(A2 ∩ A3 ).
Using the assumption of independence (implicit in the sampling with replacement setting), we have
Pr(A1 ) = (0.8)10 , Pr(A2 ) = (0.7)10 , Pr(A3 ) = (0.5)10 , and Pr(A1 ∩ A2 ) = (0.5)10 , Pr(A1 ∩ A3 ) = (0.3)10 ,
Pr(A2 ∩ A3 ) = (0.2)10 , resulting in Pr(A) = 0.1356.