Download Stat 134: exam solutions

Stat 134: exam solutions
Michael Lugo
September 28, 2011
1. [1] Consider the following events:
(i) in flipping 12 fair coins, exactly 6 heads are obtained.
(ii) in flipping 108 fair coins, either 53, 54, or 55 heads are obtained.
Which of the following statements is true?
(a) The event (i) is considerably more likely than the event (ii).
(b) These two events have approximately the same probability.
(c) The event (ii) is considerably more likely than the event (i).
Answer. The answer is (b). The two events √
have approximately the same probability.
The number of
√ heads in (i) has mean 6 and SD 3; the number of heads in (ii) has√mean
54 and SD√3 3; thus both of these quantities can be approximated by Φ((1/2)/ 3) −
Φ((−1/2)/ 3). Alternatively, the SD in (ii) is three times that in (i) by the square root law
and the interval we’re looking at is three times as wide.
46 students got this right, out of 100.
2. [1] Say P (A|B) < P (B|A). Which of the following is true?
(a) P (A) < P (B).
(b) P (A) = P (B)
(c) P (A) > P (B)
(d) There is not enough information to tell.
Answer. By the definition of conditional probability, we have P (AB)/P (B) < P (AB)/P (A).
Canceling gives 1/P (B) < 1/P (A); taking reciprocals gives P (B) > P (A). The answer is
(a).
75 students got this right, out of 100.
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3. [total 9] I toss a fair coin repeatedly, until the third time it comes up heads, and I
record the total number of coin tosses. For example, if I obtain the results T T HHT T H, I
record the number 7. What is:
(a) [2] the probability that this experiment requires exactly 7 tosses?
(b) [2] the probability of observing the exact sequence T T HHT T H, given that the experiment required seven tosses?
(c) [2] the probability that this experiment requires at least 8 tosses?
(d) [3] I toss a coin repeatedly, until the 30th time it comes up heads, and I record the
total number of coin tosses. What is the approximate probability that this requires at least
70 tosses?
Answer. (a) In this case we must have two
heads in the first six tosses, followed by a
head on the seventh toss. The probability is 62 (1/2)2 (1/2)4 (1/2) = 62 (1/2)7 = 15/128.
(b) The probability of this exact sequence is 1/128. The conditional probability is
(1/128)/(15/128) = 1/15.
(Alternatively: there are 62 outcomes which require seven tosses, of which one is this
particular sequence.)
(c) In order for the experiment to require at least eight tosses, there
must
be 7no more
7
7
7
than two heads in the first seven tosses. This has probability 0 + 1 + 2 /2 = (1 +
7 + 21)/128 = 29/128.
(d) This is the probability of obtaining at most 29 heads in 69 tosses. The number of
heads
in 69 tosses is approximately normal, with mean 69/2 = 34.5 and standard deviation
p
69 × 1/2 × 1/2 – call these µ and σ. The answer is
29.5 − µ
Φ
= Φ(−1.20) = 0.1151.
σ
Remarks. A lot of people used the wrong binomial coefficient in (a). 73 was common.
In (c), a typical error was to claim that “at least eight tosses” and “exactly seven tosses” are
complementary, and therefore that the answer is 1 minus the answer in (a). In (d), many
people took the √
probability that at least seven tosses are required to get three heads, and
then divided by 10, invoking the “square root law”; this is not correct but if you thought
of this you should be happy with yourself, as it shows some probabilistic intuition.
The average score on this question was 5.87 (out of 9), the median was 6, and the standard
deviation was 2.09.
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4. [total 5] The events A, B, and C are independent. P (AB) = 0.4, P (AC) = 0.2, and
P (ABC) = 0.1.
(a) [2] What are the probabilities of A, B, and C?
(b) [3] What is the probability that exactly one of A, B, and C occurs? (If you couldn’t
solve the previous question, give a formula for the answer to this question in terms of P(A),
P(B), and P(C).)
Answers. (a) We know P (A)P (B) = 0.4, P (A)P (C) = 0.2, P (A)P (B)P (C) = 0.1.
Dividing, P (C) = 0.25, P (B) = 0.5, P (A) = 0.8.
(b)
P (A)(1 − P (B))(1 − P (C)) + (1 − P (A))P (B)(1 − P (C)) + (1 − P (A))(1 − P (B))P (C)
= (0.8)(1 − 0.5)(1 − 0.25) + (1 − 0.8)(0.5)(1 − 0.25) + (1 − 0.8)(1 − 0.5)(0.25)
= 0.8 × 0.5 × 0.75 + 0.2 × 0.5 × 0.75 + 0.2 × 0.5 × 0.25
= 0.30 + 0.075 + 0.025 = 0.4
Remarks. There don’t seem to be any really typical errors here; mostly people lost
points just for computational reasons. The average score was 4.10 (out of 5), the median
was 5 (57 out of 100 students got full credit), and the standard deviation was 1.29.
3
5. [total 8] In a distant land, an evil zookeeper named Theo has two barrels. Barrel
number one contains three hippos and four monkeys. Barrel number two contains five hippos
and three monkeys.
(a) [2] Theo selects a barrel at random, and then removes a random animal from the
barrel. What is the probability that the animal is a hippo?
(b) [3] Theo selects a barrel at random, and then removes a random animal from the
barrel. It is a hippo. What is the probability that the animal Theo removed came from
barrel number one?
(c) [3] Theo selects a barrel at random, and then remove a random animal from the
barrel. It is a hippo. He does not put the hippo back in its barrel. Theo now removes
another animal from the same barrel. What is the probability that the second animal Theo
removes is a monkey?
Answers.
We introduce some notation: B1 and B2 are the events that each of the two barrels are
chosen. H is the event that a hippo is chosen, and M is the event that a monkey is chosen.
(a) P (H) = P (B1 H) + P (B2 H) = P (B1 )P (H|B1 ) + P (B2 )P (H|B2 ). Substituting gives
(1/2)(3/7) + (1/2)(5/8) = 59/112 ≈ 0.527.
(b) P (B1 |H) = P (B1 H)/P (H) = (1/2)(3/7)/(59/112) = 24/59.
(c) We now have two draws. Let H1 , H2 be the events that the first and second draws,
respectively, yield hippos, and let M1 and M2 be the same for monkeys. Then we compute
P (M2 |H1 ) =
P (H1 M2 B1 )P (H1 M2 B2 )
P (B1 )P (H1 M2 |B1 ) + P (B2 )P (H1 M2 |B2 )
P (H1 M2 )
=
=
P (H1 )
P (H1 )
P (H1 )
Now P (H1 M2 |B1 ) = (3/7)(4/6) and P (H1 M2 |B2 ) = (5/8)(3/7), so we get
P (M2 |H1 ) =
(1/2)(3/7)(4/6) + (1/2)(5/8)(3/7)
31
=
≈ 0.525.
59/112
59
Remark. I have friends who have a stuffed monkey named Theo. Maybe he dreams about
putting monkeys and hippos in barrels. Maybe he doesn’t dream at all, because he’s stuffed.
In general people did well on this question – in particular I expected the computation in (c)
would give people more trouble than it did! The most common errors I saw were finding
P (B1 H) instead of P (B1 |H), and P (H1 M2 ) instead of P (M2 |H1 ) – that is, confusing a
conditional probability given by a fraction with that fraction’s numerator.
The average score on this question was 6.43 (out of 8), the median was 7, and the standard
deviation was 1.71.
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6. [total 7] I roll a fair die 500 times. What is the approximate probability that I obtain
(a) [2] exactly 100 sixes?
(b) [2] at least 95 sixes?
(c) [3] I carry out this experiment 200 times. What is the approximate probability that
I roll exactly 100 sixes at least twice?
p Answers. (a) The number of sixes is binomial, with µ = 500/6 = 250/3 and σ =
500(1/6)(5/6) = 50/6 = 25/3.
The probability is Φ((100.5 − µ)/σ) − Φ((99.5 − µ)/σ) = Φ(2.06) − Φ(1.94) = 0.0065.
(b) The probability is 1 − Φ((94.5 − µ)/σ) = 1 − Φ(1.34) = 0.0901.
(c) The expected number of times getting at least 100 sixes is around (0.0065)(200) = 1.3
and the distribution is Poisson. The answer is 1 − (1 + 1.3)exp(−1.3) = 0.37. (There’s been
lots of rounding here.)
Remarks.
Fairly straightforward; most mistakes were compuational. A few people
100
wrote 500
(1/6)
(5/6)400 for (a); this doesn’t get full credit. Some people then claimed to
100
be able to evaluate the binomial coefficient. I was surprised by this – can your calculators
do that?
The average score on this question was 5.18 (out of 8), the median was 6, and the standard
deviation was 1.78.
5
7. I have three dice. One is six-sided, one is eight-sided, and one is twelve-sided. The
n-sided die has numbers 1 through n on its sides, and each side is equally likely to come up.
I roll the three dice.
(a) [2] What is the probability that I roll the same number on all three dice?
(b) [2] What is the probability that the product of the three numbers I roll is even?
(Hint: the product of two odd numbers is odd; the product of an even number with any
other number is even.)
(c) [3] What is the probability that the smallest number I roll is a 3?
(d) [3] I pick one of the dice at random, with all three equally likely to be picked. I roll
it and it comes up 4. What’s the probability the die I rolled was the six-sided die?
Answers. (a) There are six ways to roll the same number on all three dice, out of
6 × 8 × 12 equally likely outcomes. Answer: 6/(6 · 8 · 12) = 1/96.
(b) One minus the probability that the product is odd. Each die has probability 1/2 of
giving an odd number, so this is 1 − (1/2)3 = 7/8.
(c) Let M be the minimum. Then P (M = 3) = P (M ≥ 3)−P (M ≥ 4). P (M ≥ 3) is the
probability of no 1s or 2s, which is (4·6·10)/(6·8·12). Similarly P (M ≥ 4) = (3·5·9)/(6·8·12).
The answer is
240 − 135
105
4 · 6 · 10 − 3 · 5 · 9
=
=
.
6 · 8 · 12
576
576
(d) The probability of getting the six-sided die and rolling 4 is (1/3)(1/6). The probability
of rolling a 4 is (1/3)(1/6) + (1/3)(1/8) + (1/3)(1/12). The answer is the ratio of these
P (six-sided|4) =
(1/3)(1/6)
4
4
P (six-sided and 4)
=
=
= .
P (4)
(1/3)(1/6) + (1/3)(1/8) + (1/3)(1/12)
4+3+2
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Remarks. In (a), 1/(6×8×12) was common; this is the probability that we get the same
specified number on all three dice. In (b), 1/2 was common; I think people got products
confused with sums, and it is in fact the case that the sum has probability 1/2 of being even.
(This is the case as long as at least one die has an even number of sides.) In (c), a lot of
people confused the event “the smallest number is a 3” with “the largest number is a 3” or
“all the numbers rolled are smaller than 3”. Also, one answer that arose multiple times was
(1/6)(5/8)(9/12) + (3/6)(1/8)(9/12) + (3/6)(5/8)(1/12) = 87/576; this is the probability
that the smallest number rolled is a 3 and there is only one 3. (d) was for the most part
good, although I was surprised how many answers like 0.442 and 0.446 I got from people
who had just plugged everything into their calculator. This question is probably a bit easier
than it looks from the grades as I think some people didn’t get to it.
The average score on this question was 6.63 (out of 10), the median was 7, and the
standard deviation was 2.39.
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Overall remarks. As a class you did a good job on this exam. The exam was out of 41
points; the mean grade was 29.4 and the median was 31. The standard deviation was 7.3;
the quartiles were 25, 31 (median), and 35. Overall I’m pretty happy with the results of the
exam.
A more detailed grade distribution is as follows:
39-40 37-38 35-36 33-34 31-32 29-30 27-28 25-26 23-24 19-22 15-18 11-14
4
11
14
10
13
9
10
6
9
5
4
3
You may be wondering about your grade at the end of the semester. As stated at the
beginning of the semester, I intend for the distribution of grades to be approximately 30%
A, 30% B, 30% C, and 10% other; therefore it may interest you to know that 30% of the
class got 34 or better, 60% of the class got 29 or better, and 90% of the class got 19 or
better. So roughly speaking, 34 or better is an A, 29 to 34 is a B, 19 to 29 is a C, and
below 19 is a D or F. However your grade at the end of the semester will be determined from
the numerical scores you have received – I’ll calculate a weighted average of your homework
and exam grades with the weights given on the syllabus, and the people with the top thirty
percent or so of those scores will get A or A− grades, and so on.
Now stop worrying about your grade and worry about learning instead. If you learn the
material the grade will take care of itself.
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6-10
2