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Math 1332 Practice Problems for Probability
#1 How many possible ways can a five question, true-false opinion survey be answered?
#2 How many different social security numbers can there be?
#3 Joey has four shirts and two pair of shorts. How many outfits can he wear?
#4 Consider the non-biased spinner below. How many possible outcomes are there in an experiment that
consists of spinning the spinner three times and observing the number indicated by the needle?
5
1
4
2
3
#5 Consider the non-biased spinner above. Name the sample space of an experiment that consists of
spinning the spinner twice and observing the number indicated by the needle.
#6 Consider the non-biased spinner above, what is the probability of the needle indicating a three if the
spinner is spun once?
#7 If a person is chosen at random, what is the probability that he/she will have been born in March if there
is an equal likelihood of being born in any given month?
#8 If two balls are selected at random without replacement from a container holding three golf balls and two
ping-pong balls, what is the probability none are golf balls?
#9 One card is selected from a standard well-shuffled 52-card deck. Find the probability that the card is
either a heart or an ace.
#10 What is the probability of drawing an ace or a two from a standard 52-card deck of playing cards?
#11 Two cards are selected from a standard well-shuffled 52-card deck. Find the probability that both cards
are hearts.
#12 Two fair dice are cast. The numbers that fall uppermost are observed. Find the probability that a sum of
three is the result of one throw.
#13 Two fair dice are cast. numbers that fall uppermost are observed. Find the probability that a sum of
seven is the result of the throw.
#14 Two fair dice are cast. numbers that fall uppermost are observed. Find the probability that the sum is
even, given that one die is a six.
#15 Consider an experiment of randomly drawing a ball and observing the ball's color from a bag that holds
four green balls and six red balls. Construct a probability distribution for the experiment.
#16 Find the expected value of a random variable x having the following probability distribution:
–$1 –$5 –$10 $2 $100
x
.2
.1
.15 .05
P(x) .5
#17 Consider a situation where the mean number of cars pulling into the drive-thru lane of a restaurant
during a two-minute interval is two. What is the probability that the number of vehicles exceeds three?
#18 Consider a set of weather conditions such that a meteorologist can determine from past data that there
exists a 25% chance for rain each of the next four days. What is the probability, that it will rain exactly once
during the four days?
Solutions to Practice Problems for Probability
#1 How many possible ways can a five question, true-false opinion survey be answered?
There are five specific tasks (each question) and two ways to accomplish each task (true or
false), so there are 32 ways to answer the survey, 25 = 32.
#2 How many different social security numbers can there be?
A social security number has nine digits. Each digit in a social security number has ten
possible digits (0 through 9), so there are 109 possible social security numbers.
#3 Joey has four shirts and two pair of shorts. How many outfits can he wear?
There are two tasks to perform (select a shirt and a pair of shorts). There are four ways to
complete the first task of selecting a shirt and two ways to complete the second task of
selecting a pair of shorts, so there are 8 outfits, 4×2 = 8.
#4 Consider the non-biased spinner below. How many possible outcomes are there in an experiment that
consists of spinning the spinner three times and observing the number indicated by the needle?
5
There are three specific tasks (three spins) and five ways to
accomplish each task (one through five), so there are 53 or 125
outcomes to the experiment.
1
4
2
3
#5 Consider the non-biased spinner above. Name the sample space of an experiment that consists of
spinning the spinner twice and observing the number indicated by the needle.
{(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3),
(4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5)}
#6 Consider the non-biased spinner above, what is the probability of the needle indicating a three if the
spinner is spun once?
favorable outcomes 1
P ( X = 3) =
=
total outcomes
5
#7 If a person is chosen at random, what is the probability that he/she will have been born in March if there
is an equal likelihood of being born in any given month?
P ( X = March ) =
favorable outcomes month of March 1
=
=
total outcomes
12 months
12
#8 If two balls are selected at random without replacement from a container holding three golf balls and two
ping-pong balls, what is the probability none are golf balls?
Not drawing any golf balls is the same as drawing two ping-pong balls. Let event A represent
drawing a ping-pong ball on the first draw and event B represent drawing a ping-pong ball
on the second draw (assuming a ping-pong ball was drawn on the first draw). The
probability of A equals 2/5 because there are two ping-pong balls and five total balls in the
bag. The probability of B represents 1/4 because there is one ping-pong ball and four total
balls remaining. The probability of A∩B is the probability of A and B occurring:
2 1 1
P ( A ∩ B) = × =
5 4 10
#9 One card is selected from a standard well-shuffled 52-card deck. Find the probability that the card is
either a heart or an ace.
Let event A represent drawing a heart and event B represent drawing an ace. The
probability of A equals 13/52 because there are thirteen hearts and fifty-two cards. The
probability of B represents 4/52 because there are four aces and fifty-two cards. The
probability of A and B occurring, that is A∩B, equals 1/52 due to the ace of hearts. If two
events are not mutually exclusive the probability of the union of the two events, that is A or
B, is given by the formula P(A ∪ B) = P(A) + P(B) – P(A∩B). So, the probability of A ∪ B is
shown below.
P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )
13 4 1
+ −
52 52 52
16
P ( A ∪ B) =
52
4
P ( A ∪ B) =
13
P ( A ∪ B) =
#10 What is the probability of drawing an ace or a two from a standard 52-card deck of playing cards?
Let event A represent drawing an ace and event B represent drawing a two. The probability
of A equals 4/52 because there are four aces and fifty-two cards. The probability of B
represents 4/52 because there are four twos and fifty-two cards. Since A and B are mutually
exclusive, the probability of drawing an ace or a two equals 8/52 which reduces to 2/13:
P ( A ∪ B ) = P ( A) + P ( B )
4
4
P( A ∪ B ) =
+
52 52
8
P( A ∪ B ) =
52
2
P( A ∪ B ) =
13
#11 Two cards are selected from a standard well-shuffled 52-card deck. Find the probability that both cards
are hearts.
Let event A represent drawing a heart on first draw and event B represent drawing a heart
on the second draw (assuming a heart was drawn first). The probability of A equals 13/52
because there are thirteen hearts and fifty-two cards. The probability of B represents 12/51
because there are twelve hearts out of the fifty-one cards remaining. The probability of A
and B occurring equals P(A)·P(B):
13 12
⋅
52 51
1 4
P ( A) ⋅ P( B) = ⋅
4 17
1 4
P ( A) ⋅ P( B) = ⋅
4 17
1
P ( A) ⋅ P( B) =
17
P ( A) ⋅ P( B) =
#12 Two fair dice are cast. The numbers that fall uppermost are observed. Find the probability that a sum of
three is the result of one throw.
The outcomes of (2,1) and (1,2) have a sum of three, so the probability equals 2/36 which
reduces to 1/18.
#13 Two fair dice are cast. The numbers that fall uppermost are observed. Find the probability that a sum of
seven is the result of the throw.
The outcomes of (6,1), (5,2), (4,3), (3,4) , (2,5), and (1,6) have a sum of seven, so the
probability equals 6/36 which reduces to 1/6.
#14 Two fair dice are cast. The numbers that fall uppermost are observed. Find the probability that the sum
is even, given that one die is a six.
Let event A represent an even sum. Let event B represent getting a six on one die. The
probability of event A is 18/36, because there are eighteen outcomes with an even sum out of
thirty-six possible outcomes. The probability of event B occurring equals 11/36 because there
are eleven outcomes with a six on one die: (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,6), (2,6),
(3,6), (4,6), & (5,6). The probability of A∩B, the intersection of A and B, equals the number
of outcomes that have a six on one die and an even sum, which is 5/36.
P ( A given B ) =
P ( A ∩ B)
P ( B)
5 36
11 36
5 36
P ( A given B ) = ⋅
36 11
5
P ( A given B ) =
11
P ( A given B ) =
#15 Consider an experiment of randomly drawing a ball and observing the ball's color from a bag that holds
four green balls and six red balls. Construct a probability distribution for the experiment.
Let x represent the color of the ball:
x
P(x)
green
.4
red
.6
#16 Find the expected value of a random variable x having the following probability distribution:
–$1 –$5 –$10 $2 $100
x
.2
.1
.15 .05
P(x) .5
Expected Value = −1(.5) − 5(.2) − 10(.1) + 2(.15) + 100(.05)
Expected Value = −0.5 − 1 − 1 + 0.3 + 5
Expected Value = 2.8
#17 Consider a situation where the mean number of cars pulling into the drive-thru lane of a restaurant
during a two-minute interval is two. What is the probability that the number of vehicles exceeds three?
First, we calculate the probability of exactly zero cars entering the lane as shown below.
e −2 ⋅ 2 0
0!
P ( X = 0 ) ≈ 0.1353
P ( X = 0) =
Second, we calculate the probability of exactly one car entering the lane as well as exactly
two cars, and exactly three cars.
e −2 ⋅ 21
1! ,
P ( X = 1) ≈ 0.2707
P ( X = 1) =
e −2 ⋅ 22
e −2 ⋅ 23
P ( X = 3) =
2! ,
3!
P ( X = 2 ) ≈ 0.2707 P ( X = 3) ≈ 0.1804
P ( X = 2) =
Finally, to find the probability that the number of cars exceeds three, we use the Complement
Principle as below.
P ( X > 3) = 1 − ⎡⎣ P ( ( X = 0 ) ∪ ( X = 1) ∪ ( X = 2 ) ∪ ( X = 3) ) ⎤⎦
P ( X > 3) = 1 − ⎡⎣ P ( X = 0 ) + P ( X = 1) + P ( X = 2 ) + P ( X = 3) ⎤⎦
P ( X > 3) ≈ 1 − [ 0.1353 + 0.2707 + 0.2707 + 0.1804]
P ( X > 3) ≈ 1 − [0.8571]
P ( X > 3) ≈ 0.1429
#18 Consider a set of weather conditions such that a meteorologist can determine from past data that there
exists a 25% chance for rain each of the next four days. What is the probability, that it will rain exactly once
during the four days?
To answer this question, we recall Binomial Probability Formula. Let W1 represent exactly
one instance of rain in four instances of the given set of conditions. P ( W1 ) , then, represents
the probability that there is one occurrence of rain in four trials:
P ( W1 ) = C ( n, X ) ⋅ ⎡⎣ P ( R ) ⎤⎦ ⋅ ⎡⎣1 − P ( R ) ⎤⎦
X
P ( W1 ) = C ( 4,1) ⋅ ( 0.25 ) ⋅ ( 0.75 )
1
P ( W1 ) = 4 ⋅ ( 0.25 ) ⋅ ( 0.75 )
P ( W1 ) = 0.421875
3
4 −1
n− X