Download Math 511 Practice Exam 1B Solutions

Name____________________________
Math 511
Practice Exam #1B
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No Calculators
1. (a). Suppose that A, B and C are events in a sample space S and that
P(A) = 0.4, P(B) = 0.3, P(A ∪ B) = 0.5 , P(C) = 0.7
What is the value of:
(i). P(A ∪ B) = ______________________
Solution:
First note that P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.2
So, P(A ∪ B) = 1 − P(A ∩ B) = 0.8
(ii). P(A | B) = ________________________
Solution: P(A | B) =
P(A ∩ B) 0.20 2
=
= .
P(B)
0.30 3
(iii). P ( A ∩ B | A ∪ B ) = ________________________
Solution: P ( A ∩ B | A ∪ B ) =
P(A ∩ B) 0.2
=
= 0.4 .
P(A ∪ B) 0.5
(iv). P ( A | B ) = ________________________
Solution: P ( A | B ) =
P( A ∩ B)
P( B)
=
P(A) − P ( A ∩ B ) 0.4 − 0.2 0.2 2
=
=
=
1− P ( B )
1− 0.3
0.7 7
(v). If the events A and C are independent, then
P ( A ∪ C ) = ________________________
Solution: P ( A ∪ C ) = P ( A ) + P ( C ) − P ( A ∩ C ) = 0.4 + 0.70 − 0.28 = 0.82
Name____________________________
2. A box contains 10 red marbles, 10 blue marbles, 10 green marbles, 10 yellow marbles
and 10 white marbles. Suppose that 8 marbles are chosen at random and without
replacement. Then what is the probability that
(a). They consist of 4 each of two different colors?
⎛ 5⎞ ⎛ 10⎞ 2
⎜ ⎟⎜ ⎟
⎝ 2⎠ ⎝ 4 ⎠
Solution:
⎛ 50⎞
⎜ ⎟
⎝ 8⎠
(b). They consist of 3 of one color and 5 of a second color?
⎛10⎞ ⎛ 10⎞
5 × 4⎜ ⎟ ⎜ ⎟
⎝ 3 ⎠⎝ 5 ⎠
Solution:
⎛ 50⎞
⎜ ⎟
⎝ 8⎠
3. You have a fair coin and a biased coin. The biased coin comes up heads with probability
0.8 and tails with probability 0.2. You choose a coin at random and flip it.
It comes up heads. You now flip it again, what is the probability that it is again heads?
Solution: Let H1 be the event that we get a head on the first flip of the chosen coin, and let
H2 denote the event that we get a head on the second flip of the chosen coin.
P(H1 ) = P(H1 ∩ fair) + P(H1 ∩ biased) = P(H 1 | fair) P(fair) + P(H1 | biased)P(biased)
1 1 4 1 1 2 13
= ⋅ + ⋅ = + =
.
2 2 5 2 4 5 20
Now, the probability of getting two heads in two consecutive flips of the chosen coin is
P(HH) = P(HH ∩ Fair) + P(HH ∩ Biased) = P(HH | Fair)P(Fair) + P(HH | Biased)P(Biased)
2
1 1 ⎛ 4⎞ 1 1 8
89
=
= ⋅ +⎜ ⎟ ⋅ = +
.
4 2 ⎝ 5⎠ 2 8 25 200
So, P(H2 | H1 ) =
P(H1 ∩ H2 ) P(HH)
=
=
P(H1 )
P( H1 )
89
200
13
20
=
89
130
Name____________________________
4. Suppose that there are 18 songs on your iPod, and you have 7 favorites among them.
You play the songs in a random order (using the shuffle control on your iPod).
What is the probability that the 4th of your favorite songs (three have already been
played) is the 8th song played?
Solution:
Let A and B be the events:
A - the first 7 songs played consist of 3 favorites and 4 non-favorites.
B - the 8th song is a favorite.
Then we want the value of P(A ∩ B) .
⎡ ⎛ 7 ⎞ ⎛ 11⎞ ⎤
⎢
⎥
4 ⎢ ⎜⎝ 3 ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎥
.
P(A ∩ B) = P(B | A)P(A) = ⋅
11 ⎢ ⎛ 18 ⎞ ⎥
⎢ ⎜ ⎟ ⎥
⎣ ⎝ 7⎠ ⎦
5. You have to choose between two jars of candy. Jar A contains 12 pieces of chocolate
and 8 pieces of hard candy. Jar B contains 6 pieces of chocolate and 14 pieces of hard
candy. You randomly choose two pieces of candy from the first jar (without
replacement) and discover that they are both hard candies. What is the probability that
you chose the candy from Jar A? Express your answer in terms of binomial coefficients.
Solution: Let BH denote the event that both choices are hard candy. Let A denote the
event that they were chosen from Jar A, and let B denote that they were chosen from jar
B. Then, P(BH ) = P(BH ∩ A) + P(BH ∩ B) = P(BH | A)P(A) + P(BH | B)P(B) =
⎛ 8⎞
⎛ 14 ⎞
⎜⎝ 2 ⎟⎠ 1 ⎜⎝ 2 ⎟⎠ 1
⋅ +
⋅ . So,
⎛ 20 ⎞ 2 ⎛ 20 ⎞ 2
⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
⎛ 8⎞
⎜⎝ 2 ⎟⎠ 1
⋅
⎛ 20 ⎞ 2
⎛ 8⎞
⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
P(A ∩ BH ) P(BH | A)P(A)
28
P(A | BH ) =
=
=
=
=
⎛ 8⎞
⎛ 14 ⎞
⎛ 8 ⎞ ⎛ 14 ⎞ 119
P(BH )
P(BH )
⎜⎝ 2 ⎟⎠ 1 ⎜⎝ 2 ⎟⎠ 1 ⎜⎝ 2 ⎟⎠ + ⎜⎝ 2 ⎟⎠
⋅ +
⋅
⎛ 20 ⎞ 2 ⎛ 20 ⎞ 2
⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
Note : The part in blue would be acceptable for the final answer on the exam.
Name____________________________
6. (a). Suppose that 10 biased coins, each with probability 0.7 of coming up heads, are
independently flipped at random.
What is the probability that exactly 6 of them come up heads?
Solution: This is a binomial problem with n = 10, and p = 0.7. If X denotes the number
⎛ 10 ⎞
of heads, then P(X = 6) = ⎜ ⎟ (0.7)6 (0.3)4
⎝ 6⎠
(b). If each of 400 boxes contains 2 red marbles and 98 blue marbles and a marble is
chosen at random from each box, what is the probability that at least one red
marble gets picked and what is the approximate value of this probability in terms
of the constant e?
Solution: The probability that we never pick a red marble is
⎛ 98 ⎞
⎜⎝ 100 ⎟⎠
400
2 ⎞
⎛
= ⎜ 1−
⎝ 100 ⎟⎠
400
4
100
4
⎡⎛
2 ⎞ ⎤ ⎡1⎤
1
= ⎢⎜ 1−
≈⎢ 2⎥ = 8 .
⎥
⎟
e
⎣⎝ 100 ⎠ ⎦ ⎣ e ⎦
Thus the probability that we do pick at least one red marble is approximately 1−
1
.
e8
7. (a). If a lottery consists of drawing 10 positive digits ( 1 - 9) in order (with replacement
so that digits may be repeated). You choose some selection of digits with repetitions
allowed, and if some permutation of your selection agrees with the drawing, then you
win. What is the probability that some permutation of 3 3 4 4 4 6 6 6 6 6 is the winning
number?
Solution: There are 910 ways to draw the 10 digits and there are 10! 2!3!5! permutations
10!
2!3!5!
of the digits 3 3 4 4 4 6 6 6 6 6, so we get
910
(b). Suppose that 12 marbles are placed randomly into 5 boxes. What is the probability
that the first box has 4 marbles and the second is empty?
⎛ 12 ⎞
Solution: There are 512 ways to place the 12 marbles into the boxes. There are ⎜ ⎟ 38
⎝ 4⎠
ways to choose 4 marbles for the first box and then place the remaining 8 marbles into
⎛ 12 ⎞ 8
⎜⎝ 4 ⎟⎠ 3
the remaining 3 boxes. So we get,
512
Name____________________________
8. (a). Suppose that two dice are rolled. If they come up doubles, then the random variable
X is assigned the value 12, if they come up with a sum of 5, then X is assigned the value
18 and otherwise X is assigned the value 36.
What is the expected value of X?
1
1
13
Solution: E(X) = × 12 + × 18 + × 36 = 30
6
9
18
(b). What does it mean to say that “X is b(n, p)”?
Solution: X is a binomial random variable that denotes the number of successes in n
trials of independent events each having probability p of success. X has the probability
⎛ n⎞
distribution function p(k) = P(X = k) = ⎜ ⎟ p k q n − k , where q = 1 − p.
⎝ k⎠
9. What is the probability that a randomly chosen subset of {1, 2, 3, 4, 5, 6, a, b, c, d, e}
contain exactly 3 digits and any number of letters or exactly three letters and any
number of digits?
⎛ 6⎞ 5 ⎛ 5⎞ 6 ⎛ 6⎞ ⎛ 5⎞
⎜⎝ 3⎟⎠ 2 + ⎜⎝ 3⎟⎠ 2 − ⎜⎝ 3⎟⎠ ⎜⎝ 2 ⎟⎠
Solution:
211
10. Suppose that Benford’s law accurately describes the distribution of the first digits of
randomly chosen powers of 2. What is the probability that the second digit in a
randomly chosen power of 2 will be a 3 given that the initial digit is a 1? You may leave
your answer in terms of logs.
Solution:
P(second is 3 | first is 1) =
Probability the number begins with 13 log14 − log13
=
..
Probability the number begins with 1
log 2