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STAT/MA 416 Answers
Homework 3
September 13, 2007
Solutions by Mark Daniel Ward
PROBLEMS
4. Write E for the event that at least one of a pair of fair dice lands on 6. Write Fi for the
event that the sum of the dice is i (for 2 ≤ i ≤ 12). Then
P (E | Fi ) =
P (E ∩ Fi )
P (Fi )
We observe that P (E ∩ Fi ) = 0 for 2 ≤ i ≤ 6, because if at least one die lands on 6, then
the sum of the dice must be at least 7. So
P (E | Fi ) = 0
for 2 ≤ i ≤ 6
For the values of i with i ≥ 7, we use P (E | Fi ) =
P (E | F7 ) = 2/36
= 1/3
6/36
2/36
P (E | F10 ) = 3/36 = 2/3
P (E∩Fi )
P (Fi )
2/36
P (E | F8 ) = 5/36
= 2/5
2/36
P (E | F11 ) = 2/36 = 1
to compute
P (E | F9 ) = 2/36
= 1/2
4/36
1/36
P (E | F12 ) = 1/36 = 1
5. Write W for the event that the first two balls selected are white. Write B for the event
that the last two balls selected are black. Then the desired probability is P (W ∩ B) =
(6)
P (B | W )P (W ). We see that P (W ) = 152 = 1/7. Given then W occurs, then 4 white balls
(2)
6
(9)
and 9 black balls remain, so P (B | W ) = 132 = 6/13. So P (W ∩ B) = 17 13
= 6/91.
(2)
9. Note that one ball is selected from each urn. Write W1 for the event that the ball from
urn A is white. Write W2 for the event that the ball from urn B is white. Write W3 for
the event that the ball from urn C is white. The event that exactly two white balls were
selected is
(W1 ∩ W2 ∩ W3c ) ∪ (W1 ∩ W2c ∩ W3 ) ∪ (W1c ∩ W2 ∩ W3 )
Note that the events W1 ∩ W2 ∩ W3c and W1 ∩ W2c ∩ W3 and W1c ∩ W2 ∩ W3 are mutually
disjoint. The desired probability is
P W1 | (W1 ∩ W2 ∩ W3c ) ∪ (W1 ∩ W2c ∩ W3 ) ∪ (W1c ∩ W2 ∩ W3 )
Equivalently,
c
c
c
P W1 ∩ (W1 ∩ W2 ∩ W3 ) ∪ (W1 ∩ W2 ∩ W3 ) ∪ (W1 ∩ W2 ∩ W3 )
c
c
c
P (W1 ∩ W2 ∩ W3 ) ∪ (W1 ∩ W2 ∩ W3 ) ∪ (W1 ∩ W2 ∩ W3 )
1
2
Simplifying, we obtain
W3c )
W2c
P (W1 ∩ W2 ∩
∪ (W1 ∩
∩ W3 )
P (W1 ∩ W2 ∩ W3c ) ∪ (W1 ∩ W2c ∩ W3 ) ∪ (W1c ∩ W2 ∩ W3 )
Since the events W1 , W2 , W3 are independent, we compute
2
8
1
1
6
12
1
1
4
2
1
1
12
6
1
18
4
1
1
12
6
1
1
P (W1 ∩ W2 ∩ W3c ) = P (W1 )P (W2 )P (W3c ) =
P (W1 ∩ W2c ∩ W3 ) = P (W1 )P (W2c )P (W3 ) =
P (W1c ∩ W2 ∩ W3 ) = P (W1c )P (W2 )P (W3 ) =
So the desired probability is
1
1
+ 36
6
1
1
+ 36 + 19
6
3
1
4
1
1
1
4
1
1
1
4
1
= 1/6
= 1/36
= 1/9
= 7/11.
14a. We write B1 for the event that the first ball selected is black, B2 for the event that
the second ball selected is black, W3 for the event that the third ball selected is white, and
W4 for the event that the fourth ball selected is white.
(7)
Note that P (B1 ) = 121 = 7/12. Given that B1 occurs, then there are 5 white and 9
(1)
(9)
black balls in the urn. So P (B2 | B1 ) = 141 = 9/14. Given that B1 and B2 occur, then
(1)
(5)
there are 5 white and 11 black balls in the urn. So P (W3 | B1 ∩ B2 ) = 161 = 5/16. Given
(1)
that B1 and B2 and W3 occur, then there are 7 white and 11 black balls in the urn. So
(7)
P (W4 | B1 ∩ B2 ∩ W3 ) = 181 = 7/18.
(1)
So the desired probability is P (B1 ∩ B2 ∩ W3 ∩ W4 ), which can be rewritten as
7
9
5
7
P (B1 )P (B2 | B1 )P (W3 | B1 ∩B2 )P (W4 | B1 ∩B2 ∩W3 ) =
= 35/768
12
14
16
18
16. Write E for the event that a baby survives, and F for the event that the baby is delivered
by C section. Then
P (E) = P (E | F )P (F ) + P (E | F c )P (F c )
We are given P (E) = .98. We are also given P (F ) = .15, so P (F c ) = .85. We are also given
P (E | F ) = .96. Thus
.98 = (.96)(.15) + P (E | F c )(.85)
Solving for P (E | F c ) gives the desired probability of P (E | F c ) ≈ .9835.
19a. Write E for the number of people that took the class. Write M for the number of men
that took the class and W for the number of women that took the class. So E = M + W .
We are given that M = (.62)(E), and thus W = (.38)(E).
3
The total number who attended the party was (.48)(W ) + (.37)(M ), or equivalently,
(.48)(.38)(E) + (.37)(.62)(E) = (.4118)(E). The total number of women who attended
the party was (.48)(W ) = (.48)(.38)(E) = (.1824)(E). So the fraction of those attending
the party that were women is .1824E
≈ .4429; in other words, 44.29% of those attending the
.4118E
party were women.
19b. The original class had E participants. The number of participants at the party was
(as computed above) exactly (.4118)(E). So the fraction of the original class that attended
the party was .4118E
= .4118; in other words, 41.18% of the original class attended the party.
E
22a. The number of ways that B, Y, R can be distinct is exactly (6)(5)(4) = 120; to see
this, just assign any value to B, and any of the 5 remaining values of Y , and any of the 4
remaining values of R. The total number of assignments is 63 = 216. So the probability that
no two of the dice land on the same number is exactly 120
= 5/9.
216
22b. The number of ways that B, Y, R can be chosen to be distinct with B < Y < R is
exactly 63 = 20; to see this, just choose 3 distinct numbers in the range 1, . . . , 6, and assign
the smallest to B, the middle value of Y , and the largest value to R. Therefore, given that
no two of the dice land on the same number, the conditional probability that B < Y < R is
20/216
= 1/6.
exactly 120/216
Another way to get the result above is to realize that there are exactly 6 = 3! ways of
assigning the letters B, Y, R to any set of 3 distinct values in the range 1, . . . , 6. So exactly
1/6 of the results from part (a.) have the ordering B < Y < R.
20
5
22c. As above, 20 of the 216 outcomes have B < Y < R, so P (B < Y < R) = 216
= 54
.
23a. Write E for the event that the ball selected from urn II is white. Write F for the event
that the transferred ball is white; so F c is the event that the transferred ball is red. So
2
2
1
4
c
c
P (E) = P (E | F )P (F ) + P (E | F )P (F ) =
+
= 4/9
3
6
3
6
23b. Now we compute
P (E ∩ F )
P (E | F )P (F )
P (F | E) =
=
=
P (E)
P (E)
2
3
2
6
4/9
= 1/2
30. We refer to the box with 1 black and 1 white marble as “Box 1”. We refer to the box
with 2 black and 1 white marble as “Box 2”. Write Bi for the probability that the ith box
was selected. Write E for the event that the selected marble is black; so E c is the event that
the selected marble is white.
First, the probability that the selected marble is black is
1
1
2
1
P (E) = P (E | B1 )P (B1 ) + P (E | B2 )P (B2 ) =
+
= 7/12
2
2
3
2
Next, the probability that the first box was the one selected, given that the marble is
white, is
1
1
P (B1 ∩ E c )
P (E c | B1 )P (B1 )
c
2
2
P (B1 | E ) =
=
=
7 = 3/5
P (E c )
1 − P (E)
1 − 12
4
34. The only thing that changes in Example 3f is that P (C | G) = .9 (instead of “1”). So
(.9)(.6)
the probability that the suspect is guilty becomes (.9)(.6)+(.2)(.4)
= 27
≈ .8710.
31
37a. Write H1 for the event that the chosen coin shows heads on the first toss. Write F for
the event that the fair coin is selected. Then
1
1
P (F ∩ H1 )
P (H1 | F )P (F )
2
= 1/3
P (F | H1 ) =
=
= 1 1 2
1
c
c
P (H1 )
P (H1 | F )P (F ) + P (H1 | F )P (F )
+
(1)
2
2
2
37b. Write H2 for the event that the chosen coin shows heads on the first toss. Then
P (F | H1 ∩ H2 ) =
P (H1 ∩ H2 | F )P (F )
P (F ∩ H1 ∩ H2 )
=
P (H1 ∩ H2 )
P (H1 ∩ H2 | F )P (F ) + P (H1 ∩ H2 | F c )P (F c )
1
1
= 1/5
= 1 1 4 2
+ (1) 12
4
2
37c. Given that a tail appears on the third toss, then the probability that it is the twoheaded coin is 0, so the probability that it is the fair coin is 1 in this case. (A two-headed
coin will never shows tails!)
43. Write E1 for the event that the two-headed coin is selected; write E2 for the event that
the fair coin is selected; write E3 for the event that the coin which shows heads 75 percent
of the time is selected. Write H for the event that heads appears on the selected coin. Then
P (E1 | H) =
P (E1 ∩ H)
P (H | E1 )P (E1 )
=
P (H)
P (H | E1 )P (E1 ) + P (H | E2 )P (E2 ) + P (H | E3 )P (E3 )
(1) 31
= 4/9
=
(1) 13 + 12 13 + 34 13
44. Write E1 for the event that A is to be executed; write E2 for the event that B is to be
executed; write E3 for the event that C is to be executed.
Write F2 for the event that the jailer says that B will be set free; write F3 for the event
that the jailer says that C will be set free.
If A is to be executed, then it makes sense to assume that the jailer will choose to say that
B will be set free half of the time and will say that C will be set free half of the time; i.e.,
if A is to be executed, the jailer will randomly choose to say that B or C will be executed.
Then
P (E1 | F2 ) =
P (E1 ∩ F2 )
P (F2 | E1 )P (E1 )
=
P (F2 )
P (F2 | E1 )P (E1 ) + P (F2 | E2 )P (E2 ) + P (F2 | E3 )P (E3 )
1
1
2
3
= 1/3
= 1 1
1
1
+
(0)
+
(1)
2
3
3
3
Similarly, we also see that P (E1 | F3 ) = 1/3. So the prisoner is correct: The prisoner’s
probability of being executed is still 1/3, even if the jailer tells him the name of one of his
fellow prisoners who will go free. The conditional probability does not go up to 1/2.
5
47. Write E for the event that all of the balls selected are white. Write Fi for the event
that the fair die shows value i (for 1 ≤ i ≤ 6). Then
P (E) =
6
X
P (E | Fi )P (Fi )
i=1
(5i )
for all i. Therefore,
(15i )
6
6
5 5
X
X
1
2
2
1
1
1
1 1
i
i
=
=
+
+
+
+
+ 0 = 5/66
P (E) =
15
6
6 i=1 15i
6 3 21 91 273 3003
i
i=1
We note that P (Fi ) = 1/6 for all i. Also, P (E | Fi ) =
48. Write E1 for the event that a silver coin is found in the first drawer that is opened, and
E2 for the event that a silver coin is found in the second drawer that is opened. Write F1 for
the event that cabinet A is selected. Write F2 for the event that cabinet B is selected. Then
P (E2 | E1 ) =
P (E1 ∩ E2 )
P (E1 ∩ E2 | F1 )P (F1 ) + P (E1 ∩ E2 | F2 )P (F2 )
=
P (E1 )
P (E1 | F1 )P (F1 ) + P (E1 | F2 )P (F2 )
1
(1) 2 + (0) 21
= 2/3
=
(1) 12 + 21 12
51a. Write E for the event that she receives a new job offer. Write F1 for the event that her
recommendation is strong; write F2 for the event that her recommendation is good; write F3
for the event that her recommendation is weak. Then
P (E) = P (E | F1 )P (F1 ) + P (E | F2 )P (F2 ) + P (E | F3 )P (F3 )
= (.80)(.7) + (.40)(.2) + (.10)(.1) = .65
51b. Given that she receives the offer, the conditional probability that she received a strong
recommendation is
P (E ∩ F1 )
P (E | F1 )P (F1 )
(.80)(.7)
P (F1 | E) =
=
=
= 56/65
P (E)
P (E)
.65
Similarly, the conditional probability that she received a good recommendation is
P (F2 | E) =
P (E | F2 )P (F2 )
(.40)(.2)
P (E ∩ F2 )
=
=
= 8/65
P (E)
P (E)
.65
Similarly, the conditional probability that she received a weak recommendation is
P (F3 | E) =
P (E ∩ F3 )
P (E | F3 )P (F3 )
(.10)(.1)
=
=
= 1/65
P (E)
P (E)
.65
51c. Given that she does not receive the offer, the conditional probability that she received
a strong recommendation is
P (F1 | E c ) =
P (E c ∩ F1 )
P (E c | F1 )P (F1 )
(.20)(.7)
=
=
= 2/5
c
P (E )
1 − P (E)
1 − .65
Similarly, the conditional probability that she received a good recommendation is
P (F2 | E c ) =
P (E c ∩ F2 )
P (E c | F2 )P (F2 )
(.60)(.2)
=
=
= 12/35
c
P (E )
1 − P (E)
1 − .65
6
Similarly, the conditional probability that she received a weak recommendation is
P (F3 | E c ) =
P (E c ∩ F3 )
P (E c | F3 )P (F3 )
(.90)(.1)
=
=
= 9/35
c
P (E )
1 − P (E)
1 − .65
53. Write E for the event that system 1 is functioning; write F for the event that the system
is functioning. Note that P (F ) = 1 − P (F c ) = 1 − (1/2)n , since F c occurs (i.e., the system
is not functioning) if and only if all n of the components fail. So
P (E | F ) =
P (E ∩ F )
P (F | E)P (E)
(1)(1/2)
=
=
P (F )
P (F )
1 − (1/2)n
Multiplying by 2n in the numerator and denominator yields P (E | F ) =
2n−1
.
2n −1
56. Fix n throughout the problem. Write E for the event that the nth coupon is of
S a new
type. Write Fi for the event that the nth coupon is of type i. We note that S = m
i=1 Fi ,
and the Fi ’s are mutually disjoint. The probability that the nth coupon is a new type is
P (E) =
m
X
P (E ∩ Fi )
i=1
Note that E ∩ Fi occurs if each of the first n − 1 types are not type of i, and the nth coupon
is of type i. So P (E ∩ Fi ) = (1 − pi )n−1 pi . So the probability that the nth coupon is a new
type is
m
X
P (E) =
(1 − pi )n−1 pi
i=1
65a. Write E for the event that the couple gives the correct answer; write F for the event
that the couple agrees.
There are four outcomes:
1. Write G1 for the event that both people are correct. (Note P (G1 ) = (.6)(.6) = .36.)
2. Write G2 for the event that only the wife is correct. (Note P (G2 ) = (.4)(.6) = .24.)
3. Write G3 for the event that only the husband is correct. (Note P (G3 ) = (.6)(.4) = .24.)
4. Write G4 for the event that neither person is correct. (Note P (G4 ) = (.4)(.4) = .16.)
Note that the Gi ’s are mutually disjoint. So
P (E | F ) =
P (E ∩ F )
P (G1 )
P (G1 )
.36
=
=
=
= 9/13
P (F )
P (G1 ∪ G4 )
P (G1 ) + P (G4 )
.36 + .16
65b. Given that the couple disagrees, then they flip a coin. If the outcome of the coin
corresponds to the person with the correct answer (either the wife or the husband, but not
both!), then the couple has the correct answer. So the conditional probability is 1/2. (The
outcome simply depends on the outcome of the coin!)
67c. There are exactly nj distinct ways to pick j out of the n components; for each such
selection, the probability that the j chosen components work and the other n−j components
n−j
each fail is exactly P
pj (1 − p)
. So the probability that at least k of the n components are
working is exactly nj=k nj pj (1 − p)n−j .
7
70. Write Ei for the event that the ith child has hemophilia. Write F for the event that
the queen is a carrier. Then
P ((E1c ∩ E2c ∩ E3c ) ∩ F )
P (E1c ∩ E2c ∩ E3c )
P (E1c ∩ E2c ∩ E3c | F )P (F )
=
P (E1c ∩ E2c ∩ E3c | F )P (F ) + P (E1c ∩ E2c ∩ E3c | F c )P (F c )
(1/2)3 (1/2)
=
= 1/9
(1/2)3 (1/2) + (1)(1/2)
P (F | E1c ∩ E2c ∩ E3c ) =
If there is a fourth prince, the probability that he will have hemophilia is
P ((E1c ∩ E2c ∩ E3c ) ∩ E4 )
P (E1c ∩ E2c ∩ E3c )
P ((E1c ∩ E2c ∩ E3c ) ∩ E4 | F )P (F ) + P ((E1c ∩ E2c ∩ E3c ) ∩ E4 | F c )P (F c )
=
P (E1c ∩ E2c ∩ E3c | F )P (F ) + P (E1c ∩ E2c ∩ E3c | F c )P (F c )
(1/2)4 (1/2) + (0)(1/2)
=
= 1/18
(1/2)3 (1/2) + (1)(1/2)
P (E4 | E1c ∩ E2c ∩ E3c ) =
73. In all five parts below, there are 25 = 32 possible outcomes.
73a. There are two possibilities, namely, all are boys or all are girls. So the probability is
2/32 = 1/16.
73b. There is only one possibility, namely, the 3 eldest are boys and the others (i.e., the
youngest two) are girls.
So the probability is 1/32.
73c. There are 53 = 10 distinct ways to select three of the five children as the boys; the
other two children will be girls in each case. So the probability is 10/32 = 5/16.
73d. The two oldest children are girls, and there are 23 = 8 remaining possibilities for the
others (i.e., the three youngest children). So the probability is 8/32 = 1/4.
73e. There is only way to have all boys, so 31 of the 32 ways have at least one girl. So the
probability is 31/32.
74. Write En for the event that A’s nth roll is the last roll, i.e., A does not roll a “9” on
her first n − 1 tries, and B does not roll
a “6” on5 her
n−1first4 n − 132tries,
n−1 and
A does roll a “9”
4 n−1
31 n−1 4
on her nth try. Then P (En ) = 1 − 36
1 − 36
= 36
.
36
36
36
Notice that the En ’s are disjoint, and the probability that the final roll is made by A is
!
n−1 n−1 ∞
∞
∞ [
X
X
32
31
4
4
1
9
5 =
En =
P (En ) =
=
P
4
36
36
36
36 1 − 36 36
19
n=1
n=1
n=1
Another method is to use conditional probabilities. On a particular move, we note that A
and B both get their desired totals (“9” for A and “6” for B) with probability (4/36)(5/36);
in such a case, A wins. On a particular move, we note that only A (but not B) gets her
desired total (“9” for A, but not “6” for B) with probability (4/36)(31/36); in such a case, A
wins. On a particular move, we note that only B (but not A) gets her desired total (not “9”
for A, but “6” for B) with probability (32/36)(5/36); in such a case, B wins. So, given that
8
the game is ending, the desired probability is
(4/36)(5/36) + (4/36)(31/36)
= 9/19
(4/36)(5/36) + (4/36)(31/36) + (32/36)(5/36)
77a. Write E1 for the event that the first trial results in outcome 1; write F for the event
that outcome 3 is the last of the three outcomes to occur. Then
P (E1 | F ) =
P (E1 ∩ F )
P (F | E1 )P (E1 )
=
P (F )
P (F )
We know that P (E1 ) = 1/3 since all three outcomes are equally likely to appear. Also,
P (F ) = 1/3. Finally, P (F | E1 ) = 1/2; to see this, notice that we are given the appearance
of outcome 1 on the first roll, so the next new outcome to appear sometime in the future is
equally likely to be outcome 2 or 3. Summarizing these results yields
P (E1 | F ) =
(1/2)(1/3)
= 1/2
1/3
77b. Write E2 for the event that the second trial results in outcome 1. Then
P (E1 ∩ E2 | F ) =
P ((E1 ∩ E2 ) ∩ F )
P (F | (E1 ∩ E2 ))P (E1 ∩ E2 )
=
P (F )
P (F )
We know that P (E1 ∩ E2 ) = 1/9 since all three outcomes are equally likely to appear each
time, and the trials are independent. Also, P (F ) = 1/3. Finally, P (F | (E1 ∩ E2 )) = 1/2; to
see this, notice that we are given the appearance of outcome 1 on the first two rolls, so the
next new outcome to appear sometime in the future is equally likely to be outcome 2 or 3.
Summarizing these results yields
P (E1 ∩ E2 | F ) =
(1/2)(1/9)
= 1/6
1/3
84a. Write E for the event that A wins. Write En for the event that A wins on her nth
draw from the urn. When the balls are replaced after each draw, then the probability of En
3(n−1) 1 4 3(n−1) 4
8 n−1 1
is exactly 1 − 12
= 32
= 27
. The En ’s are disjoint, so the
12
3
3
probability that the final roll is made by A is
!
n−1 ∞
∞
∞ [
X
X
8
1
1
1
P
En =
P (En ) =
=
8 = 9/19
27
3
3
1
−
27
n=1
n=1
n=1
Similarly, write F for the event that B wins. Write Fn for the event that B wins on her
nth draw from the urn. When the balls are replaced after each draw, then the probability
4
3(n−1) 2 1 4
8 n−1 2
4 3(n−1)
of Fn is exactly 1 − 12
1 − 12
= 23
= 27
. The Fn ’s are
12
3
3
9
disjoint, so the probability that the final roll is made by B is
!
n−1 ∞
∞
∞ [
X
X
8
2
2
1
P
Fn =
P (Fn ) =
=
8 = 6/19
27
9
9 1 − 27
n=1
n=1
n=1
Similarly, write G for the event that C wins. Write Gn for the event that C wins on her nth
draw from the urn. When the balls are replaced after each draw, then the probability of Gn
9
4
2 1
4
4
2 3(n−1) 2
4 3(n−1)
1 − 12
1 − 12
=
=
is exactly 1 − 12
12
3
3
3
3
Gn ’s are disjoint, so the probability that the final roll is made by C is
P
∞
[
!
Gn
n=1
8 n−1
27
4
27
. The
∞
X
n−1 ∞ X
8
4
4
1
=
P (Gn ) =
=
8 = 4/19
27
27
27 1 − 27
n=1
n=1
A different method of solution is available by conditional probabilities. Write a “round” as a
sequence of three turns, by A, then B, then C. There are 33 = 27 types of rounds possible;
on 27 − 23 = 19 of these types of rounds, at least one player draws a white ball. On 9 of
these 19 types, player A gets a white ball, so the probability that A wins is 9/19. On 6 of
these 19 types, player A does not get a white ball, but player B does get a white ball, so
the probability that B wins is 6/19. On 4 of these 19 types, neither player A nor player B
get a white ball, but player C does get a white ball, so the probability that C wins is 4/19.
84b. If the withdrawn balls are not replaced, then the probability that A wins is
4
+
P (E) =
12
8
12
7
11
6
10
4
8
7
6
5
4
3
4
7
+
=
9
12
11
10
9
8
7
6
15
The probability that B wins is
P (F ) =
4
8
7
6
5
4
+
11
12
11
10
9
8
8
53
7
6
5
4
3
2
4
+
=
12
11
10
9
8
7
6
5
165
8
12
The probability that C wins is
P (G) =
7
4
8
7
6
5
4
4
+
11
10
12
11
10
9
8
7
8
7
6
5
4
3
2
1
4
7
+
=
12
11
10
9
8
7
6
5
4
33
8
12
85a. The answers are exactly the same as above for part (a.). In other words, the prob9 6 4
abilities that players A, B, C win are 19
, 19 , 19 , respectively. Just notice that each player is
always drawing from an urn containing 4 white balls and 8 other balls. The setup does not
change the probabilities at all, in the model with replacement.
10
85b. If the withdrawn balls are not replaced, then the probability that A wins is exactly
given below (the jth term is the probability that A wins on her jth draw, for 1 ≤ j ≤ 9)
4
+
12
3 3 3 3 3 3 8
4
8
7
4
8
7
6
4
+
+
12
11
12
11
10
12
11
10
9
3 3 3 3 8
7
6
5
4
+
12
11
10
9
8
3 3 3 3 3 7
6
5
4
4
8
+
12
11
10
9
8
7
3 3 3 3 3 3 8
7
6
5
4
3
4
+
12
11
10
9
8
7
6
3 3 3 3 3 3 3 8
7
6
5
4
3
2
4
+
12
11
10
9
8
7
6
5
3 3 3 3 3 3 3 3 8
7
6
5
4
3
2
1
4
6476548
+
=
12
11
10
9
8
7
6
5
4
13476375
The probability that B wins is exactly given below (the jth term is the probability that B
wins on her jth draw, for 1 ≤ j ≤ 8)
8
12
3 3 3 4
8
7
4
8
7
6
4
+
+
12
12
11
11
12
11
10
10
3 3 3 7
6
5
4
8
+
12
11
10
9
9
3 3 3 3 8
7
6
5
4
4
+
12
11
10
9
8
8
3 3 3 3 3 8
7
6
5
4
3
4
+
12
11
10
9
8
7
7
3 3 3 3 3 3 8
7
6
5
4
3
2
4
+
12
11
10
9
8
7
6
6
3 3 3 3 3 3 3 8
6994
7
6
5
4
3
2
1
4
+
=
12
11
10
9
8
7
6
5
5
22275
11
The probability that C wins is exactly given below (the jth term is the probability that C
wins on her jth draw, for 1 ≤ j ≤ 8)
2 3 2 3 3 2 8
4
8
7
4
8
7
6
4
+
+
12
12
12
11
11
12
11
10
10
3 3 3 2 8
7
6
5
4
+
12
11
10
9
9
3 3 3 3 2 8
7
6
5
4
4
+
12
11
10
9
8
8
3 3 3 3 3 2 7
6
5
4
3
4
8
+
12
11
10
9
8
7
7
3 3 3 3 3 3 2 8
7
6
5
4
3
2
4
+
12
11
10
9
8
7
6
6
3 3 3 3 3 3 3 2 8
7
6
5
4
3
2
1
4
922819
+
=
12
11
10
9
8
7
6
5
5
4492125
THEORETICAL EXERCISES
11. Write En for the event of tossing at least one head in the first n tosses of a coin, with
probability p of landing heads each time. So
P (En ) = 1 − P (Enc ) = 1 − (1 − p)n
We want P (En ) ≥ 1/2, i.e., 1 − (1 − p)n ≥ 1/2, or equivalently, 12 ≥ (1 − p)n . Taking the
natural logarithm of both sides, ln(1/2) ≥ ln((1 − p)n ), or equivalently, ln(1/2) ≥ n ln(1 − p).
ln(1/2)
Dividing by ln(1 − p) (which is a negative number), we obtain ln(1−p)
≤ n.