INTRODUCTION TO PROBABILITY (REVIEW OF
HOMEWORK QUESTIONS FROM JIM PITMAN'S
PROBABILITY TEXTBOOK.) QUESTIONS &
ANSWERS RATED 100% CORRECT!!
1.1.4 Equally Likely Outcomes
Suppose I bet on red at roulette and you bet on black, both bets on the same
spin of the wheel.
a) What is the probability that we both lose?
b) What is the probability that at least one of us wins?
c) What is the probability that at least one of us loses? Answer - a) 2/38
because only 2 options out of 38 are neither black or red.
b) 36/38 this is the opposite of a
c) 100% at least 1 player always loses.
1.1.5
Suppose a deck of 52 cards is shuffled and the top two cards are dealt.
a) How many ordered pairs of cards could result in an outcome?
,b) Assuming each of the pairs has an equal chance, calculate the chance the
first card is an ace
c) the chance the second card is an ace (explain your answer by symmetry
argument as well as by counting)
d) the chance that both cards are aces
e) the probability that at least one is an ace Answer - a) 52 x 51 = 2652, you
have to subtract by one the second time because the cards are not replaced.
b) 4x51 = 204 pairs with first card ace, then 204/2652 = 1/13
c) because the outcomes are equally likely, it doesn't matter if it comes first or
second. Therefore by the symmetry rule it is exactly the same as if it were first
1/13.
d) 4x3= 12 pairs both aces. 12/2652 = 1/221
e) P(first card ace) +P(second card ace) - P(intersection)
1/13 + 1/13 -1/221 = 33/221
1.1.6
Repeat exercise 5, supposing instead that after the first card is dealt, it is
replaced, and shuffled into the deck before the second card is dealt.
,a) How many ordered pairs of cards could result in an outcome?
b) Assuming each of the pairs has an equal chance, calculate the chance the
first card is an ace
c) the chance the second card is an ace (explain your answer by symmetry
argument as well as by counting)
d) the chance that both cards are aces
e) the probability that at least one is an ace Answer - a) 52 x 52 =2704
b) 4x52 same
c) 4x52 same
d) 4x4 = 16, 16/2704=1/169
e) 1/13 + 1/13 -1/169 = 25/169
1.1.7
Suppose two dice are rolled. Find the probability of the following events.
a) The maximum of the two numbers rolled is less than or equal to 2
b) maximum of the two numbers rolled is less than or equal to 3.
, c) the maximum of the two numbers is exactly equal to 3
d) repeat b & c for x instead of 3, for each x from one to 6
e) Denote P(x) the probability that the maximum number is exactly x. What
should P1+p2+p3+p4+p5+p6 equal? Answer - a) P(number rolled is <= 2) =
(2x2)/(6x6) = 1/9, There are only two options for first and second dice, and you
find the total options by squaring the options of one.
b) P(number rolled is <= 3)
(3x3)/36 = 1/4
c) P(the maximum rolled is exactly 3)
(1,3)
(3,1)
(2,3)
(3,2)
(3,3)
=5/36
d) P(max number rolled is <=x) = x^2/36
P(max number rolled is exactly x) = (2x-1)/36, for this one draw a table
e. (1/36) + (3/36) + (5/36) + (7/36) + (9/36) + (11/36) = 1
1.1.8
Repeat exercise 7 for two rolls of a n-sided die for an arbitrary n instead of 6.
HOMEWORK QUESTIONS FROM JIM PITMAN'S
PROBABILITY TEXTBOOK.) QUESTIONS &
ANSWERS RATED 100% CORRECT!!
1.1.4 Equally Likely Outcomes
Suppose I bet on red at roulette and you bet on black, both bets on the same
spin of the wheel.
a) What is the probability that we both lose?
b) What is the probability that at least one of us wins?
c) What is the probability that at least one of us loses? Answer - a) 2/38
because only 2 options out of 38 are neither black or red.
b) 36/38 this is the opposite of a
c) 100% at least 1 player always loses.
1.1.5
Suppose a deck of 52 cards is shuffled and the top two cards are dealt.
a) How many ordered pairs of cards could result in an outcome?
,b) Assuming each of the pairs has an equal chance, calculate the chance the
first card is an ace
c) the chance the second card is an ace (explain your answer by symmetry
argument as well as by counting)
d) the chance that both cards are aces
e) the probability that at least one is an ace Answer - a) 52 x 51 = 2652, you
have to subtract by one the second time because the cards are not replaced.
b) 4x51 = 204 pairs with first card ace, then 204/2652 = 1/13
c) because the outcomes are equally likely, it doesn't matter if it comes first or
second. Therefore by the symmetry rule it is exactly the same as if it were first
1/13.
d) 4x3= 12 pairs both aces. 12/2652 = 1/221
e) P(first card ace) +P(second card ace) - P(intersection)
1/13 + 1/13 -1/221 = 33/221
1.1.6
Repeat exercise 5, supposing instead that after the first card is dealt, it is
replaced, and shuffled into the deck before the second card is dealt.
,a) How many ordered pairs of cards could result in an outcome?
b) Assuming each of the pairs has an equal chance, calculate the chance the
first card is an ace
c) the chance the second card is an ace (explain your answer by symmetry
argument as well as by counting)
d) the chance that both cards are aces
e) the probability that at least one is an ace Answer - a) 52 x 52 =2704
b) 4x52 same
c) 4x52 same
d) 4x4 = 16, 16/2704=1/169
e) 1/13 + 1/13 -1/169 = 25/169
1.1.7
Suppose two dice are rolled. Find the probability of the following events.
a) The maximum of the two numbers rolled is less than or equal to 2
b) maximum of the two numbers rolled is less than or equal to 3.
, c) the maximum of the two numbers is exactly equal to 3
d) repeat b & c for x instead of 3, for each x from one to 6
e) Denote P(x) the probability that the maximum number is exactly x. What
should P1+p2+p3+p4+p5+p6 equal? Answer - a) P(number rolled is <= 2) =
(2x2)/(6x6) = 1/9, There are only two options for first and second dice, and you
find the total options by squaring the options of one.
b) P(number rolled is <= 3)
(3x3)/36 = 1/4
c) P(the maximum rolled is exactly 3)
(1,3)
(3,1)
(2,3)
(3,2)
(3,3)
=5/36
d) P(max number rolled is <=x) = x^2/36
P(max number rolled is exactly x) = (2x-1)/36, for this one draw a table
e. (1/36) + (3/36) + (5/36) + (7/36) + (9/36) + (11/36) = 1
1.1.8
Repeat exercise 7 for two rolls of a n-sided die for an arbitrary n instead of 6.
