Download MDM4U Probability Test 17

MDM 4U Data Management and Statistics
Probability Test 17
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Multiple Choice [12 K/U]
Identify the choice that best completes the statement or answers the
question.
1. A coin is tossed three times. What is the probability of tossing three
heads in succession?
3
1
1
7
a.
b.
c.
d.
8
2
8
8
Answer: c.
2. A coin is tossed three times. What is the probability of tossing exactly
two heads?
5
1
1
3
a.
b.
c.
d.
8
2
4
8
Answer: d.
3. If the probability of rain tomorrow is 65%, what is the probability that
it will not rain tomorrow?
a. 0.65
b. 0.55
c. 0.45
d. 0.35
Answer: d.
4. Two identical spinners each have five equal sectors that are numbered
1 to 5. What is the probability of a total of 7 when you spin both three
spinners?
4
7
1
1
a.
b.
c.
d.
25
25
4
5
Answer: a.
5. If the odds in favour of snow tomorrow are 4:7, what is the
probability of snow tomorrow?
4
7
4
7
a.
b.
c.
d.
7
11
11
4
Answer: c.
6. If you were to toss three coins, what are the odds against them landing
heads up?
a. 1:8
b. 8:1
c. 7:8
d. 7:1
Answer: d.
7. What are the odds against randomly picking a green apple from a fruit
bowl containing seven green apples and four red ones?
a. 7:4
b. 4:11
c. 7:11
d. 4:7
Answer: d.
8. Suppose you simultaneously roll a standard die and spin a spinner that
is divied into 10 equal sectors, numbered 1 to 10. What is the probability
of getting 4 on both the die and the spinner?
1
1
1
1
a.
b.
c.
d.
4
16
15
60
Answer: d.
9. Suppose you simultaneously roll a standard die and spin a spinner
with 8 equal sectors, numbered 1 to 8. What is the probability of both
rolling an even number and spinning an odd number?
1
1
1
1
a.
b.
c.
d.
8
16
4
48
Answer: c.
10. If Mike does his mathematics homework today, the probability that
he will do it tomorrow is 0.8. The probability that he will do his
homework today is 7.0. What are the odds that he will do it both today
and tomorrow?
a. 4:1
b. 8:7
Answer: d.
c. 3:2
d. 14:11
11. When two cards are drawn from a standard deck, what is the
probability of drawing a face card or an ace?
4
2
3
5
a.
b.
c.
d.
13
13
13
13
Answer: a.
12. What is the probability of randomly selecting either a club or a nonface card from a standard deck of cards?
1
1
43
10
a.
b.
c.
d.
5
4
52
13
Answer: c.
Show your work [12 A]
13. A standard die is rolled. What is the probability of rolling a prime
number? [2 A]
Answer:
1
2
14. If the odds against the Blue Jays winning this year’s World Series
are 20:1, what is the probability that the Blue Jays will win this series?
[2 A]
Answer:
1
21
15. What are the odds in favour of a total greater than 9 in a given roll of
two standard dice? [2A]
Answer: 1 : 5
16. If a satellite launch has a 97% chance of success, what is the
probability of three consecutive successful launches? [2A]
Answer: (0.7)3 ≈ 91.3%
17. The probability that Sergey will play golf today is 60%, the
probability that he will play golf tomorrow is 75%, and the probability
that he will play golf on both days is 50%. What is the probability that
he does not play golf on either day? [2A]
Solution
Event A = “Sergey will play golf today”, P (A) = 0.6.
Event B = “Sergey will play golf tomorrow”, P (B) = 0.75.
Event A∩B = “Sergey will play golf on both days”, P (A∩B) = 0.5.
P (A∪B) = P (A) + P (B) – P (A∩B) = 0.6 + 0.75 – 0.5 = 0.85.
̅̅̅̅̅̅̅
P (𝐴̅ ∩ 𝐵̅) = P (𝐴
∪ 𝐵 ) = 1 – P (A∪B) =1 – 0.85 = 0.15.
18. If 28% of the population of Statsville wears contact lenses, 37%
have blue eyes, and 9% are blue-eyed people who wear contact lenses,
what is the probability that a randomly selected resident has neither blue
eyes nor contact lenses? [2A]
Solution
Event A = “a randomly selected resident wears contact lenses”
Event B = “a randomly selected resident has blue eyes”
A∩B = “a randomly selected resident wears contact lenses and has blue
eyes”
P (A) = 0.28, P (B) = 0.37, P (A∩B) = 0.09,
P (A∪B) = P (A) + P (B) – P (A∩B) = 0.28 + 0.37 – 0.09 = 0.56.
̅̅̅̅̅̅̅
P (𝐴̅ ∩ 𝐵̅) = P (𝐴
∪ 𝐵 ) = 1 – P (A∪B) =1 – 0.56 = 0.44.
Communication
[6C]
19. Explain the difference between the empirical probability, theoretical
probability, and subjective probability. [2C]
a) Empirical probability
Answer
The empirical probability, also known as relative frequency, or
experimental probability, is the ratio of the number of outcomes in
which a specified event occurs to the total number of trials, not in a
theoretical sample space but in an actual experiment. For example, Olga
tossed a coin 100 times and got 53 tails and 47 heads. The empirical
probability of getting a tail equals 0.53.
b) Theoretical probability
Answer
Theoretical Probability of an event is the ratio of the number of
favorable outcomes to the total number of equally likely outcomes. For
example, probability of getting an even number when rolling a die
equals
3 1
P (even) = =
6 2
since the total number of equally likely outcomes equals 6 and the
number of favorable outcomes equals 3.
c) Subjective probability
Answer
A probability derived from an individual's personal judgment about
whether a specific outcome is likely to occur. Subjective probabilities
contain no formal calculations and only reflect the subject's opinions and
past experience. For example, Abdul states that the probability that he
will pass this Probability Test equals 60%.
20. Explain the meaning of the terms in the probability formula, P (A)
𝑛 (𝐴)
=
. [2C]
𝑛 (𝑆)
Answer
P (A) – the probability of an event A
𝑛 (𝐴)
𝑛 (𝑆)
– the ratio of numbers of favourable for the event A outcomes and
outcomes in the sample space S.
21. The probability that Jacqueline will be elected to the student’s
council is 0.6, and the probability that she will be selected to represent
her school in a public-speaking contest is 0.75. The probability of
Jacqueline achieving both of these goals is 0.5. [2C]
a) Are these two goals mutually exclusive? Explain your answer.
Answer No, not mutually exclusive. She can be elected to both of these
goals.
b) What is the probability that Jacqueline is either elected to the
student’s council or picked for the public-speaking contest?
Note: "either p or q" means "either p or q, but not both"
Answer:
P (either council or contest) =
P (council) + P (contest) – 2 P (council and contest) =
= 0.6 + 0.75 – 2(0.5) = 0.35.
Thinking [10T]
22. The Royals’ coach stated that “the odds in favour of us winning the
next game are 5:7, the odds of tying the next game are 1:3, and the odds
of losing the next game are 2:3”. Can the coach’s predictions be correct?
Justify your answer. [4T]
Answer:
The odds in favour of us winning the next game are 5:7, therefore,
5
5
P (winning) =
=
5 + 7 12
The odds of tying the next game are 1:3, therefore,
1
1
P (tying) =
=
1+3 4
The odds of losing the next game are 2:3, therefore,
2
2
P (losing) =
=
2+3
5
P (winning) + P (tying) + P (losing) =
=
5
1
+ +
2
12 4 5
25+15+24
64
60
16
= = ≠ 1.
60 15
Therefore, the coach’s predictions are not correct.
23. A test for the presence of E. coli in water detects the bacteria 97% of
the time when the bacteria is present, but also gives a false positive 2%
of the time, wrongly indicating the presence of E. coli in uninfected
water. If 10% of the water samples tested contain E. coli, what is the
probability that a test result indicating the presence of the bacteria is
accurate? [4 T]
Solution
Event E = “the bacteria E. coli is present in water”.
Event D = “The test detects the bacteria”.
Given: P (D|E) = 0.97, P (D|𝐸̅ ) = 0.02, P (E) = 0.1
Find: P (E|D)
P (𝐸̅ ) = 1 – P (E) = 1 – 0.1 = 0.9
P (D) = P (DE) + P (D𝐸̅ ) = P (D|E) P (E) + P (D|𝐸̅ ) P (𝐸̅ )
= (0.97) (0.1) + (0.02) (0.9)
= 0.115
(0.1)(0.97)
𝑃(𝐷𝐸)
𝑃(𝐸)𝑃(𝐷|𝐸)
𝑃(𝐸 |𝐷) =
=
=
≈ 0.84.
𝑃(𝐷)
𝑃(𝐷)
0.115
24. A study on the effects that listening to loud music through
headphones had on teenager’s hearing found that 12% of those teenagers
in the sample who did listen to music in this way showed signs of
hearing problems. If 60% of the sample reported that they listened to
loud music on headphones regularly, and 85% of the sample were found
not to have hearing problems, are the events {having hearing problems}
and {listening to loud music on headphones} independent? Explain your
reasoning. [3T]
Answer
Event A = {having hearing problems},
Event B = {listening to loud music on headphones}
P (A) = 100% – 85% = 15%, P (B) = 60%
P (A|B) = 12%
P (A|B) ≠ P (A) ⟹ the events A and B are dependent.