7-7 Conditional Probability
1. CLUBS The Spanish Club is having a potluck lunch where each student brings in a cultural dish. The 10 students
randomly draw cards numbered with consecutive integers from 1 to 10. Students who draw odd numbers will
bring main dishes. Students who draw even numbers will bring desserts. If Cynthia is bringing a dessert, what is the
probability that she drew the number 10?
SOLUTION:
Let P(A) = the probability that the number is even.
Let P(A and B) = the probability that the number is both 10 and even.
There are 10 available integers.
The sample space for event A contains 5 outcomes: (2, 4, 6, 8, 10}.
So, P(A) =
or
.
The sample space for P(A and B) contains 1 outcome: {10}.
So, P(A and B) =
.
ANSWER:
or 20%
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7-7 Conditional Probability
2. A card is randomly drawn from a standard deck of 52 cards. What is the probability that the card is a king of
diamonds, given that the card drawn is a king?
SOLUTION:
Let P(A) = the probability that the card is a king.
Let P(A and B) = the probability that the card is a king and a diamond.
There are 52 possible cards.
The sample space for event A contains 4 outcomes:{king of hearts, king of diamonds, king of spades, king of
clubs}
So, P(A) =
or
.
The sample space for P(A and B) contains 1 outcome: {king of diamonds}.
So, P(A and B) =
.
ANSWER:
or 25%
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7-7 Conditional Probability
3. GAME In a game, a spinner with the 7 colors of the rainbow is spun. Find the probability that the color spun is
blue, given the color is one of the three primary colors: red, yellow, or blue.
SOLUTION:
Let P(A) = the probability that the color spun is a primary color.
Let P(A and B) = the probability that the color spun is blue.
There are 7 possible colors.
The sample space for event A contains 4 outcomes:{red, orange, yellow, green, blue, indigo, violet}
So, P(A) =
.
The sample space for P(A and B) contains 1 outcome: {blue}.
So, P(A and B) =
.
ANSWER:
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7-7 Conditional Probability
4. Fifteen cards numbered 1–15 are placed in a hat. What is the probability that the card has a multiple of 3 on it,
given that the card picked is an odd number?
SOLUTION:
Let P(A) = the probability that the number is odd.
Let P(A and B) = the probability that the number is odd and a multiple of three.
There are 15 possible numbers.
The sample space for event A contains 8 outcomes:{1, 3, 5, 7, 9, 11, 13, 15}
So, P(A) =
.
The sample space for P(A and B) contains 3 outcomes: {3, 9, 15}.
So, P(A and B) =
.
ANSWER:
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7-7 Conditional Probability
5. A blue marble is selected at random from a bag of 3 red and 9 blue marbles and not replaced. What is the
probability that a second marble selected will be blue?
SOLUTION:
Let P(A) = the probability that the first marble selected is blue.
Let P(B) = the probability that the second marble selected is blue.
Let P(A and B) = the probability that the first and second marbles are both blue.
Since there are a total of 12 marbles to begin with, 9 of which are blue, P(A) =
.
Since the first marble was blue, now there are 3 red and 8 blue marbles in the sample space, P(B) =
, or about
73%.
Another way to look at this is P(A and B) =
.
So, the probability that a second marble selected will be blue is
or about 73%.
ANSWER:
or about 73%
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7-7 Conditional Probability
6. A die is rolled. If the number rolled is less than 5, what is the probability that it is the number 2?
SOLUTION:
Let P(A) = the probability that the number is less than 5.
Let P(A and B) = the probability that the number is less than 5 and 2.
There are 6 possible numbers.
The sample space for event A contains 4 outcomes:{1, 2, 3, 4}
So, P(A) =
.
The sample space for P(A and B) contains 1 outcome: {2}.
So, P(A and B) =
.
So, the probability that the number is 2 if the number rolled is less than 5 is
or 25%.
ANSWER:
or 25%
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7-7 Conditional Probability
7. If two dice are rolled, what is the probability that the sum of the faces is 4, given that the first die rolled is odd?
SOLUTION:
Let P(A) = the probability that the first die rolled is odd.
Let P(A and B) = the probability that the first die rolled is odd and the sum of the faces is 4.
There are 36 possible combinations of dice rolls.
The sample space for event A contains 18 outcomes:{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3,
3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}.
So, P(A) =
.
The sample space for P(A and B) contains 2 outcomes: {(1, 3), (3, 1)}.
So, P(A and B) =
.
So, the probability that the sum of the faces is 4, given that the first die rolled is odd is
or about 11%.
ANSWER:
or about 11%
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7-7 Conditional Probability
8. A spinner numbered 1 through 12 is spun. Find the probability that the number spun is an 11 given that the number
spun was an odd number.
SOLUTION:
Let P(A) = the probability that the number spun is odd.
Let P(A and B) = the probability that the number spun is odd and 11.
There are 12 possible numbers.
The sample space for event A contains 6 outcomes:{1, 3, 5, 7, 9, 11}
So, P(A) =
.
The sample space for P(A and B) contains 1 outcome: {11}.
So, P(A and B) =
.
So, the probability of spinning a number that is 11 given that it is an odd numbers is
or about 17%.
ANSWER:
or about 17%
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7-7 Conditional Probability
9. If two dice are rolled, what is the probability that the sum of the faces is 8, given that the first die rolled is even?
SOLUTION:
Let P(A) = the probability that the first die rolled is even.
Let P(A and B) = the probability that the first die rolled is even and the sum of the faces is 8.
There are 36 possible combinations of dice rolls.
The sample space for event A contains 18 outcomes:{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4,
3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
So, P(A) =
.
The sample space for P(A and B) contains 3 outcomes: {(2, 6), (4, 4), (6, 2)}.
So, P(A and B) =
.
So, the probability that the sum of the faces is 8, given that the first die rolled is even is
or about 17%.
ANSWER:
or about 17%
10. PICNIC A school picnic offers students hamburgers, hot dogs, chips, and a drink.
a. At the picnic, 60% of the students order a hamburger and 48% of the students order a hamburger and chips.
What is the conditional probability that a student who orders a hamburger also orders chips?
b. If 50% of the students ordered chips, are the events of ordering a hamburger and ordering chips independent?
Explain.
c. If 80% of the students who ordered a hot dog also ordered a drink and 35% of all the students ordered a hot
dog, find the probability that a student at the picnic orders a hot dog and drink. Explain.
SOLUTION:
Let P(A) = the probability that student orders a hamburger.
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7-7 Conditional Probability
Let P(A and B) = the probability that student orders a hamburger and chips.
So, P(A) =
.
The sample space for P(A and B) =
So, the conditional probability is 0.8 or 80%.
b. No; P(chips) = 0.5 and P(chips | hamburger) = 0.8. Because these two probabilities are not the same, the
events are dependent.
c. 28%; P(drink | hot dog) = 0.8 and P(hot dog) = 0.35. Because P(drink | hot dog) =
,
P(drink and hot dog) = P(drink | hot dog) • P(hot dog) = (0.8)(0.35) = 0.28.
ANSWER:
a. 0.8 or 80%
b. No; P(chips) = 0.5 and P(chips | hamburger) = 0.8. Because these two probabilities are not the same, the
events are dependent.
c. 28%; P(drink | hot dog) = 0.8 and P(hot dog) = 0.35. Because P(drink | hot dog) =
,
P(drink and hot dog) = P(drink | hot dog) • P(hot dog) = (0.8)(0.35) = 0.28.
11. The Venn diagram shows students’ favorite places to study, the library (L) or home (H).
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7-7 Conditional Probability
a. A total of 60 students responded to the survey. Determine the number of students who replied that they study
neither at the
library nor at home.
b. What is the probability that if a student selected the library, he or she selected the library and at home? Explain.
c. A student says that selecting the library and selecting at home are independent events. Do you agree? Explain.
SOLUTION:
a. Subtract the sum of the students who study in the library, the students who study at home, and the students who
study in either the library or at home from the total number of students who responded to the survey, 60, to find the
number of students who prefer neither the library nor at home. 60 – (12 + 6 + 36) = 60 – 54 = 6 students
b.
Let P(L) = the probability that the student studies in the library.
Let P(L and H) = the probability that the students studies in the library and at home.
There are 60 possible outcomes/students.
The sample space for event L contains 12 + 6, or 18 outcomes.
So, P(L) =
.
The sample space for P(L and H) contains 6 outcomes.
So, P(L and H) =
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7-7 Conditional Probability
c. No; If P(H|L) is the same as P(H) or P(L|H) is the same as P(L), then H and L are independent.
Because P(H|L) is
and P(H) is
, H and L are not independent.
ANSWER:
a. 6 students
b. P(H|L) =
, P(L) =
and P(H and L) =
. So, P(H|L) =
c. No; sample answer: If P(H|L) is the same as P(H) or P(L|H) is the same as P(L), then H and L are
independent.
Because P(H|L) is
and P(H) is
, H and L are not independent.
12. WRITE Let A represent the event of owning a house and let B represent the event of owning a car. Are these
events independent or dependent? How do you think P(A|B) compares to P(B|A)? Explain your reasoning.
SOLUTION:
It can expected that the probability of someone having a car, given that they have a house, would be higher than the
probability of someone having a car. This would mean that P(car | house) ≠ P(car), so the events would have to be
dependent.
It can be expected that there is a higher probability of someone owning a car if they already own a house as
compared to owning a house if they already own a car because many people can afford a car but not a house. This
means that P(B|A) > P(A|B).
ANSWER:
Sample answer: I expect the probability of someone having a car, given that they have a house, to be higher than
the probability of someone having a car. This would mean that P(car | house) ≠ P(car), so the events would have
to be dependent. I think that there is a higher probability of someone owning a car if they already own a house as
compared to owning a house if they already own a car because many people can afford a car but not a house. This
means that P(B|A) > P(A|B).
13. PERSEVERE Of all the students at North High School, 25% are enrolled in Algebra and 20% are enrolled in
Algebra and Health.
a. If a student is enrolled in Algebra, find the probability that the student is enrolled in Health as well.
b. If 50% of the students are enrolled in Health, are being enrolled in Algebra and being enrolled in Health
independent events? Explain.
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7-7 Conditional Probability
c. Of all the students, 20% are enrolled in Accounting and 5% are enrolled in Accounting and Spanish. If being
enrolled in Accounting and being enrolled in Spanish are independent events, what percent of students are enrolled
in Spanish? Explain.
SOLUTION:
a.
P(Algebra) = 0.25
P(Algebra and Health) = 0.2
So, the probability that the students is enrolled in Health, given that the student is enrolled in Algebra is 80%.
b. No; P(Health|Algebra) = 0.8 and P(Health) = 0.5. Because the two probabilities are not equal, the two
events are dependent.
c. 25%;
P(Accounting) = 0.2
P(Accounting and Spanish) = 0.05
Because the events are independent, P(Spanish|Accounting) = P(Spanish) = 0.25.
ANSWER:
a. 80%
b. No; P(Health|Algebra) = 0.8 and P(Health) = 0.5. Because the two probabilities are not equal, the two
events are dependent.
c. 25%;
P(Accounting) = 0.2
P(Accounting and Spanish) = 0.05
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7-7 Conditional Probability
Because the events are independent, P(Spanish|Accounting) = P(Spanish) = 0.25.
14. ANALYZE In a standard deck of playing cards, the face-value cards are the cards numbered 2–10. Two cards
are to be randomly drawn without replacing the first card. Find the probability of drawing two face-value cards
and the conditional probability that exactly one of those cards is a 4. Explain.
SOLUTION:
The total number of combinations of 2 cards is 52C2.
There are 36 face-value cards to the total number of combinations of 2 face-value cards is 36C2.
So, the probability of drawing two face-value cards is
. There are four cards with a 4 on them and 32 other
face-value cards left, so the number of combinations with exactly one 4 is 4 • 32C1 = 4 • 32 = 128. (See below for
calculation of this combination). The probability is
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7-7 Conditional Probability
ANSWER:
; The total number of combinations of 2 cards is 52C2 = 1326. There are 36 face-value cards so the
total number of combinations of 2 face-value cards is 36C2 = 630. So, the probability of drawing two face-value
cards is
. There are four cards with a 4 on them and 32 other face-value cards left, so the number of
combinations with exactly one 4 is 4 • 32C1 = 4 • 32 = 128. The probability is
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.
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