PH 116C, Solution set 8
1. Read Boas, Ch. 15
2. Boas, p. 724, prob. 15.1-4
bigskip A single card is drawn at random from a shuffled deck. What is the probability
that it is red? That it is the ace of hearts? That it is either a 3 or a 5? That it is
either an ace or red or both?
• the probability that one card is red is 26/52 = 1/2, as there are 26 red cards
(diamonds and hearts) out of 52 cards.
• the probability that it is the ace of hearts is 1/52, as there is only one such card.
• the probability that it is either a 3 or a 5 is 8/52 = 2/13, as there are 8 such
cards in total (four 3’s and four 5’s) out of 52.
• the probability that it is an ace or a red or both is 28/52 = 7/13, as there are 26
red cards (including 2 red aces) and 2 black aces.
3. Boas, p. 728, prob. 15.2-13
A student claims in Problem 15.1-5 that if one child is a girl, the probability that both
are girls is 21 . Use appropriate sample spaces to show what is wrong with the following
argument: It doesn’t matter whether the girl is the older child or the younger; in either
case the probability is 12 that the other child is a girl.
The sample space that describes the two children consists of four possible outcomes:
BB, BG, GB, GG where the order indicates which one is the younger and which one
is older and B/G stands for boy/girl. If one child is a girl, the reduced sample space
consists only of three possibilities: BG, GB, GG. Without any further information,
all three possibilities are equally likely. Hence, the probability of two girls is P = 13 .
The student who claims that the probability is P = 12 is wrong as long as there is
no further information. But suppose one is told that one of the girl’s names is Mary.
Now, if we denote the girl whose name is Mary by G∗ , then the reduced sample space
actually now contains four equally likely possibilities: BG∗ , G∗ B, G∗ G, GG∗ , in which
case the probability of two girls is indeed P = 21 as the student claimed. So, one must
be extremely careful on how the problem is posed.
4. Boas, p. 728, prob. 15.2-15
Use the sample spaces specified in eqs. (2.4) and (2.5) on p. 727–728 of Boas to answer
the following questions about a toss of two dice.
In each case, we will provide the answer that first employs the uniform sample space
and then employs the non-uniform sample space.
(a) What is the probability that the sum is ≥ 4?
1
• Only three events (1, 1), (1, 2) and (2, 1) give a sum less than 4, which means
that 33 events give a sum ≥ 4. Hence, the probability is
P =
33
11
=
36
12
(1)
1
2
• The associated probabilities with the sums 2 and 3 are 36
and 36
, respectively.
1
Adding up these probabilities yields 12 . Thus, the associated probabilities
with the sums ≥ 4 must add up to:
P =1−
1
11
=
12
12
(2)
as expected.
(b) What is the probability that the sum is even?
• This happens when both numbers are even or both are odd:
P =
1
1
· 18 =
36
2
(3)
• This happens for the points in the sample space where the sum is even:
P =
1
3
5
5
3
2
1
1
+
+
+
+
+
+
=
36 36 36 36 36 36 36
2
(4)
(c) What is the probability that the sum is divisible by 3?
• This happens for 12 possible combinations:
P =
1
12
=
36
3
(5)
• This happens when the sum is 3, 6, 9, or 12:
P =
2
5
4
1
1
+
+
+
=
36 36 36 36
3
(6)
(d) If the sum is odd, what is the probability that it is equal to 7?
• Our sample space is given by half of the original sample space, and we here
have 6 possible combinations
P =
6
1
=
18
3
(7)
• The modified non-uniform sample space is the one in eq. (2.5) on p. 727 of
Boas, but with new probabilities:
P (3) =
2
4
6
4
2
, P (5) =
, P (7) =
, P (9) =
, P (11) =
.
18
18
18
18
18
The probability for 7 is then
P (7) =
2
6
1
=
18
3
(8)
(e) What is the probability that the product of the numbers on the two dice is 12?
• This happens for (3, 4), (4, 3), (2, 6), (6, 2); the probability is then
P =
4
1
=
36
9
(9)
• the non-uniform sample space given by eq. (2.5) on p. 727 of Boas is not
useful in this case.
5. Boas, p. 777, prob. 15.11-9
One box contain one die and another box contains two dice. You select a box at
random and take out and toss whatever is in it (that is, toss both dice if you have
picked box 2). Let x =number of 3’s showing.
(a) Set up the sample space and associated probabilities for x.
We shall denote an element of the sample space Ω as an ordered pair of numbers. If
box 1 is chosen, then the corresponding element of the sample space is an ordered
pair of two numbers: the first number is 1 (as this was the selected box) and the
second number indicates the number shown on the first die after it is tossed. If box
2 is chosen, then the corresponding element of the sample space is an ordered pair of
three numbers: the first number is 2 (as this was the selected box), the second number
indicates the number shown on the first die after it is tossed and the third number
indicates the number shown on the second die after it is tossed.
Using this notation, the elements of the sample space are:
Ω = {(1, 1), (1, 2), . . . , (1, 6) ; (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2), . . . , (2, 6, 6)} .
|
{z
} |
{z
}
six elements
36 elements
Since there is a probability of 21 of choosing box 1 (and similarly for box 2), the
probabilities of the events above are:
• each of the events {(1, 1), (1, 2), . . . , (1, 6)} has a probability of
1
2
·
1
6
=
1
12
,
• each of the events {(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2), . . . , (2, 6, 6)} has a proba1
1
bility of 12 · 36
= 72
.
The total probability must add up to 1 and it does since
6 · 61 + 36 ·
1
72
= 1.
Let x be the number of 3’s showing. Thus the possible values of x are 0 , 1 , or 2. There
is only one elements of Ω with two 3’s showing, namely (2, 3, 3) which has a probability
of
1
.
P (x = 2) = 72
The number of elements of Ω with one 3 showing include (1, 3) with probability
and the ten elements,
1
12
(2, 1, 3) , (2, 2, 3) , (2, 4, 3) , (2, 5, 3) , (2, 6, 3) , (2, 3, 1) , (2, 3, 2) , (2, 3, 4) , (2, 3, 5) , (2, 3, 6) ,
3
each with probability
1
.
72
Hence the total probability for x = 1 is
P (x = 1) =
1
12
+ 10 ·
1
72
= 29 .
Finally, the probability of x = 0 is obtained by subtracting from 1 the probability of
x = 1 or x = 2,
1
P (x = 0) = 1 − 72
− 29 = 55
.
72
(b) What is the probability of at least one 3?
This is given by the sum of the probabilities for x = 1 and x = 2. Using the results of
part (a), we find
1
17
2
=
.
(10)
P (x ≥ 1) = +
9 72
72
(c) If at least one 3 turns up, what is the probability that you picked up the first box?
If at least one 3 turns up, there is only one element of Ω that corresponds to choosing
1
box 1, namely (1, 3) and this has probability 12
. This is a relative probability since
we are now restricted to those elements of Ω in which at least one 3 turns up. In part
17
(b), we found that the probability for x ≥ 1 is 72
. Hence, we conclude that if at least
one 3 turns up, then the probability you picked up the first box is the ratio of these
two probabilities,
P [(1, 3)]
1/12
6
=
=
.
(11)
P (x ≥ 1)
17/72
17
This result can also be understood as the conditional probability for choosing box 1
given that at least one 3 turns up. Indeed, eq. (11) is nothing more than an application
of Bayes’ formula [cf. eq. (3.8) on p. 732 of Boas].
6. Boas, p. 734, prob. 15.3-6
A card is selected from a shuffled deck. What is the probability that it is either a king
or a club? That it is both a king and a club?
The probability of having either a king (event A: 4 cards out of 52) or a club (event
B: 13 cards out of 52) is given by
P (A + B) = P (A) + P (B) − P (A ∩ B) =
1
1
1
4
+ −
=
13 4 52
13
(12)
where we have taken into account the fact that the two events are not mutually
exclusive (there is one king of clubs).
In the second case, we are looking for a king of clubs; there is only one such card, so
that the probability of picking that card is
1
52
4
(13)
7. Boas, p. 734, prob. 15.3-9
Two cards are drawn at random from a shuffled deck and laid aside without being
examined. Then a third card is drawn. Show that the probability that the third card
is spade is 14 just as it was for the first card.
Let us consider all the mutually exclusive possibilities (remembering that there are 13
spades out of 52 cards):
• if the two discarded cards are spades (probability = 14 · 12
: for the first discarded
51
card, there 13 spades out of 52 cards, while for the second we have 12 spades left
out of 51 cards), the third could be spades or something else: the probabilities of
it being a spade is
11
Ps =
(14)
50
as there are 11 spades left in a deck of 50 cards. The probability of this event is
then
11
1 4 11
·
=
;
(15)
P2 = ·
4 17 50
50 · 17
• if only the first discarded card is a spade (probability =
Ps =
12
,
50
P11 =
1
4
· (1 −
12
,
50
P12 =
), we have
1 13 12
13 · 3
·
·
=
;
4 17 50
50 · 17
• if only the second discarded card is a spade (probability =
Ps =
12
)
51
3
4
(16)
·
13
),
51
we have
3 13 12
3 · 13
·
·
=
;
4 51 50
50 · 17
(17)
Incidentally, we note that this is the same as in (16), as one expects.
• if none of the two discarded cards are spades (probability = 34 (1 − 13
) ), we have
51
Ps =
13
,
50
P0 =
3 38 13
1 13 · 19
·
·
=
.
4 51 50
2 50 · 17
(18)
The probability for the event given by the sum of these events happening is given by
the sum of their probabilities :
1
P = P2 + P11 + P12 + P0 = .
4
(19)
as one expects, because taking two cards out of the deck does not add any information
about the state of the deck. In fact, the question is the same as asking what is the
probability that the third card in the full deck is a spade; this is 1/4, as for any other
suit, or for any other position inside the deck.
8. Boas, p. 736, prob. 15.3-19
Suppose it is known that 1% of the population have a certain type of cancer. It is also
known that a test for this kind of cancer is positive for 99% of the people who have it
5
but it is also positive in 2% of the people who do not have it. What is the probability
that a person who tests positive has cancer of this type?
We are going to use Bayes formula: for P (B|A), the conditional probability of B to
happen once A has happened, we have
P (B|A) =
P (A ∩ B)
,
P (A)
(20)
P (A|B) =
P (A ∩ B)
.
P (B)
(21)
while for P (A|B), it is
Dividing these two formulas and recalling that P (B) = P (B|A)P (A)+P (B|Ac )P (Ac ),
we have
P (A|B) =
P (B|A)P (A)
P (B|A)P (A)
=
.
P (B)
P (B) = P (B|A)P (A) + P (B|Ac )P (Ac )
(22)
Now, let A be the condition in which the patient has the cancer and B be the test
being positive: the requested probability is the probability of A given that B occurs,
P (A|B) =
0.99 · 0.01
1
= .
0.99 · 0.01 + 0.02 · 0.99
3
(23)
That is, there is a 2/3 probability that the result of the test is a false positive.
9. Boas, p. 743, prob. 15.4-9
Two cards are drawn at random from a shuffled deck. What is the probability that
both are red? If at least one is red, what is the probability that both are red? If at
least one is a red ace, what is the probability that both are red? If exactly one is a
red ace, what is the probability that both are red?
The probability of both cards being red is
Prr =
26 25
25
·
=
52 51
102
(24)
because there are 26 red cards out 52 when we pick the first card, and 25 out of 51 for
the second.
For the following questions, we will make use of the conditional probability: the probability of the event B happening, provided that A happened, is1
P (B|A) =
1
In the notation of Boas, PA (B) = P (AB)/P (A).
6
P (B ∩ A)
P (A)
(25)
If at least one is red, the probability that both are red is
r
Prr
=
26
(1
52
−
26 25
·
52 51
25
) · 2 + 26
51
52
·
25
51
=
25
,
77
(26)
where in the denominator we have the probability of picking at least a red card (i.e.,
either one or two red cards), and in the numerator we have the probability of having
two red cards.
If at least one is a red ace, the probability that both are red is
a
Prr
=
2
52
· 25
+ 24
· 2
51
52 51
2
1
2
(1 − 51
) · 2 + 52
·
52
1
51
=
49
98
2 · (100 + 1) 101
(27)
where in the denominator we have the probability of picking at least a red ace (2 cards
out of 52), that is, one or two red aces, and in the numerator we have the probability
of having a red ace and a red card.
If exactly one is a red ace, the probability that both are red is
a
Prr
=
2
· 24 + 24
· 2
52 51
52 51
2
1
(1 − 51
)·2
52
=
96
12
=
2 · 100
25
(28)
where in the denominator we have the probability of picking one red ace (2 cards out
of 52), and in the numerator we have the probability of having a red ace and a red
card which is not a red ace.
10. Boas, p. 743, prob. 15.4-10
What is the probability that you and a friend have different birthdays? (For simplicity,
let a year have 365 days) What is the probability that three people have different
birthdays? Show that the probability that n people have n different birthdays is
1
2
3
n−1
p= 1−
1−
1−
... 1 −
(29)
365
365
365
365
Estimate this for n 365 by calculating ln p. Find the smallest n for which p < 12 .
Hence show that for a group of 23 people or more, the probability is greater than 12
that two of them have the same birthday.
The probability that two people have different birthdays is given by
1
p2 = 1 −
365
(30)
as this is the opposite event of having the same birthday, which has probability 1/365.
If there are three people, the probability is given by the probability that two of them
have different birthdays, multiplied by the probability that the third has another
birthday:
1
2
p3 = 1 −
1−
(31)
365
365
7
where we have 2/365 as there are 2 days out of 365 which already are birthdays. Given
n people, we can prove that the formula (29) is right: this can be proven by induction.
The cases n = 2, 3 have already been proved, so let us see that if the formula holds
for n − 1 people then it will hold also for n.
The probability that n people have different birthdays is given by the probability
that n − 1 people have different birthdays times the probability that the n-th one has
another different birthday:
n−1
1
2
3
n−1
pn = pn−1 · 1 −
= 1−
1−
1−
... 1 −
,
365
365
365
365
365
where in the first passage we have n−1
because n − 1 days out of 365 are already
365
occupied, and in the last step we have used the inductive hypothesis.
n
We can expand this expression in terms of 365
, for small n. First, let us compute the
logarithm of p, and use the property ln abc . . . = ln a + ln b + ln c + . . ., as well as the
expansion ln(1 + x) = x + O(x2 ):
n−1
ln pn ' −
1
2
3
n−1
1 X
1 n(n − 1)
−
−
− ... −
=−
k=−
365 365 365
365
365 k=1
365
2
(32)
where in the last step we have summed the arithmetic series. We note that ln pn < 0
as expected since pn < 1. We can write the probability as
1 n(n − 1)
(33)
pn ≈ exp −
365
2
We can solve to see when pn > 12 :
1 n(n − 1)
> ln 2,
365
2
=⇒
n2 − n − 730 ln 2 > 0 ,
(34)
which implies that
n < n1 or n > n2 ,
with n1,2 =
1±
√
1 + 4 · 730 ln 2
n1 = −21.99994
=
(35)
n2 = 22.99994
2
Since only positive integer values of n are relevant, we conclude that pn > 12 for n ≥ 23.
This tells us that in a group of 23 people there a 50% chance to have two people
sharing the same birthday. Note that our approximation of ln(1 + x) = x + O(x2 ) ' x
n 2
) < 1%.
is reliable, since the correction to this result is of order ( 365
11. Boas, p. 744, prob. 15.4-18
Find the number of ways of putting 3 particles in 5 boxes according to the three kind
of statistics.
• for particles obeying the Maxwell-Boltzmann statistics, the number of ways we
can distribute the 3 particles in 5 boxes is 53 ; the first particle can go in any of the
8
5 boxes and then the second one can as well go in any 5 of them, independently
from the first choice. The same happen for the third particle, so that we have
5 · 5 · 5 = 53 = 125
(36)
ways of putting 3 particles in 5 boxes.
If the particles are identical, we cannot distinguish configurations in which the
same number of particles is in any couple of boxes; for example, if we have one
particle in the first three boxes, this would correspond to six configurations in
the previous case and to one configuration in the case of identical particles. For
identical particles, we can have 2 different statistics:
• for particles obeying the Fermi-Dirac statistics, there cannot be more than one
particle per box. The number of arrangements of 3 identical particles in 5 boxes,
5!
with no more than one particle per box is C(5, 3) = 3!·2!
= 10;
• for particles obeying the Bose-Einstein statistics, we can have multiple particles
7!
per box; this means that we have C(7, 3) = 4!3!
= 35 ways.
12. Boas, p. 749, prob. 15.5-8
Would you pay $10 per throw of two dice if you were to receive a number of dollars
equal to the product of the numbers on the dice?
The random variable we are interested in is the product of the two numbers on the
dice; if its expectation value is more than 10, the game is favorable. The sample space
is the one created by the toss of two dice:
Product 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 30 36
1
2
2
3
2
4
2
1
2
4
2
1
2
2
2
1
2
1
Prob.
36 36 36 36 36 36 36 36 36 36 36 36 36 36 36 36 36 36
(37)
Then, the expectation value is
X
1
x¯ =
xi pi = (1 + 4 + 6 + 12 + 10 + 24 + 16 + 9 + 20 + 48 + 30 + 16 +
36
i
441
= 12.25
(38)
36
As the expectation value is bigger than 10, the game is favorable. In average, you will
earn $2.25 at each game.
+36 + 40 + 48 + 25 + 60 + 36) =
13. Boas, p. 755, prob. 15.6-8
Given that a particle is inside a sphere of radius 1, and that it has equal probabilities
of being found in any two volume elements of the same size, find the cumulative
distribution function F (r) for the spherical coordinate r and form it find the density
function f (r). Find r¯ and σ.
F (r) is the probability that the particle is inside a sphere of radius r (then, we have
F (1) = 1). If the particle has equal probabilities of being in any two volumes, the
probability of it being found in a region of volume V will be proportional to the volume
of that region. A sphere of radius r has a volume of 34 πr3 , hence
4
F (r) = c · V = c πr3
3
9
(39)
And we can find c in such a way that F (1) = 1; this gives c = 1/(4π/3) and
F (r) = r3
(40)
The density function f (r) is then
f (r) =
d
F (r) = 3r2
dr
(41)
The mean value of the position of the particle and the standard deviation are
Z 1
Z 1
Z 1
3 2
9
3
2
2
rf (r)dr =
3r dr = , σ = E(r ) − r¯ =
r¯ =
r2 f (r)dr −
(42)
4
16
0
0
0
Z 1
3
9
3
9
= −
= .
=
3r4 dr −
16
5 16
80
0
R
1
Alternatively, one could also have calculated σ 2 = 0 (r − r¯)2 f (r)dr, and would have
found the same result. The standard deviation is then
r
3
σ=
.
(43)
80
10