Download 6.5

Section 6.5
This section will address the idea that the sum of all probabilities is equal to one;
in other words, there is a 100% chance that one of the outcomes will occur. Also
the Complement rule, which means that if I know (or can calculate) all the
outcomes that I do not want, then if I subtract that probability from 1 that I will
have the probability of what I want. And lastly, the Additive Law, which is
𝑃[𝐴 ∪ 𝐵] = 𝑃[𝐴] + 𝑃[𝐵] − 𝑃[𝐴 ∩ 𝐵]
Problem 88-98, must be understood accomplished based on a few “rules”. First,
the capital letters represent events and the lower case letters represent simple
events (or outcomes)…In each case, the events or simple events shown in a given
problem are all that there are (none are omitted).
Problem #88
Since we have 3 simple events shown, then these 3 simple events must add up to
1…therefore, P[x] + P[y] + P[z] = 1 = 0.10 + 0.25 +P[z]
1 = 0.35 + P[z], or P[z] = 0.65
Problem #89
Since we are talking about events in this case, we must consider either the
Additive Law or a Venn Diagram…looking at the parts that are given to us in the
problem I do not have enough of the components of the Additive Law to be of any
use, so I will try a Venn Diagram. First I note that there are only two events, so it
is a two circle Venn Diagram:
A
B
Next, I note that all of A (both segments of the circle) add up to 0.35 because it
says “P[A] = 0.35”…
0.35
A
B
Next, it says “P[A∩B] = 0.18”, which means that the intersection of A and B is
0.18…
A
0.35
B
0.18
At last the “question”… “P[A∩Bc]”…this means that it is in A (so it is one of the
two segments of the circle A, but it is not in the circle B…so it is where I placed an
“x” (marks the spot).
A
0.35
x
B
0.18
Therefore, x + 0.18 must equal 0.35, so x must equal 0.35 – 0.18 = 0.17
Problem #90
Since we have 3 simple events shown, then these 3 simple events must add up to
1…therefore, P[r] + P[s] + P[t] = 1 = 0.28 +P[s] + 0.77
1 = 1.05 + P[s], or P[s] = -0.05…oops! you cannot have
a negative probability (probability must between 0 and 1), so this is not possible
Problem #91
Since we have 3 simple events shown, then these 3 simple events must add up to
1…therefore, P[x] + P[y] + P[z] = 1 = 0.55 + 0.45 +P[z]
1 = 1.0 + P[z], or P[z] = 0 (and this is fine because it is between 0 and 1)
Problem #92
Since we are talking about events in this case, we must consider either the
Additive Law or a Venn Diagram … let’s first look at the Additive Law, which is:
𝑃[𝑀 ∪ 𝑁] = 𝑃[𝑀] + 𝑃[𝑁] − 𝑃[𝑀 ∩ 𝑁]
We only have 2 of the four elements, so it cannot be used to solve this problem.
Now Venn Diagram… there are only two events, so it is a two circle Venn Diagram:
M
N
Next, I note that all of M (both segments of the circle) add up to 0.35 because it
says “P[M] = 0.6”…
M
A
0.60
N
B
Next, it says “P[MUN] = 0.8”, which means that the union of M and N is 0.8…
which means that everything in both the circles (all three segments) adds up to
0.8…what that does say is that anything outside the circles is 0.2, since everything
must add to 1; as well, the segment that is not in circle M is worth 0.2, because
0.6 + 0.2 = 0.8…
M
A
0.60
N
B
0.20
0.20
At last the “question”… “P[N]”…this means that it is both sections in the N circle.
Unfortunately, there is no way with the information given to calculate the other
segment of the N circle, so the answer is “not possible”
Problem #93
Since we have 1 simple event shown, and that’s all, and it must add up to 1… then
P[t] = 1.00
Problem #94
Since we have 4 simple events shown, then these 4 simple events must add up to
1 … therefore, P[a] + P[b] + P[c] + P[d] = 1 = 0.65 + 0.20 + 0.05 +P[d]
1 = 0.90 + P[d], or P[d] = 0.10
Problem #95
Since we given P[A] and are asked for its complement (P[Ac]) then we know these
two items MUST add to 1 (per the complement rule):
1 = 0.19 + P[Ac], or P[Ac] = 0.81
Problem #96
Since we are talking about events in this case, we must consider either the
Additive Law or a Venn Diagram … let’s first look at the Additive Law, which is:
𝑃[𝐽 ∪ 𝐾] = 𝑃[𝐽] + 𝑃[𝐾] − 𝑃[𝐽 ∩ 𝐾]
Since we are given the complement of 𝑃[𝐽 ∪ 𝐾] (𝑃[𝐽 ∪ 𝐾]c) then we know what
𝑃[𝐽 ∪ 𝐾] is since they must add to equal 1(per the complement rule)…so
𝑃[𝐽 ∪ 𝐾] = 0.4 . Therefore the equation becomes:
0.4 = 𝑃[𝐽] + 𝑃[𝐾] − 𝑃[𝐽 ∩ 𝐾]
Now add in the values for the intersection (𝑃[𝐽 ∩ 𝐾]) and the circle K (𝑃[𝐾]), and
we get: 0.4 = 𝑃[𝐽] + 0.3 − 0.1 or 0.4 = 𝑃[𝐽] + 0.2 or 𝟎. 𝟐 = 𝑷[𝑱]
Now the Venn Diagram… there are only two events, so it is a two circle Venn
Diagram:
J
K
Next, I note that all of M (both segments of the circle) add up to 0.35 because it
says “P[K] = 0.3”…
K
A
0.30
J
B
Next, it says “P[J∩K] = 0.1”, which means that the intersection of A and B is 0.1…
K
J
A 0.1
0.30
It also means that the other segment of the K circle must be 0.2, since the 2
segments must add up to be 0.30.
0.30
K
J
A
0.20 0.1
Lastly before the solution…, it says “P[JUK]c = 0.6”, which means that the
complement of the union of J and K is 0.6… which means that everything outside
both the circles equals 0.6
0.30
K
J
M
N
0.20 0.1
x
K
B
A
0.60
So all those segments including the unknown segment “x” must add up to equal
1… 0.20 + 0.10 + 0.60 + x = 1, therefore x =0.1 and that mean that the question of
what P[J] is equal to the two segments in the J circle or 0.10 + 0.10 = 0.2 = P[J]
Problem #97
I would like to rewrite the probabilities in a table (which I believe will make it a bit
easier to interpret):
Probability
a
0.10
b
0.15
c
0.15
d
0.15
e
0.20
f
0.20
g
(a) You can easily note that all the probabilities for all the simple events are here
except for “g”; and since that is what is first asked for, let’s solve it based on the
idea that all those probabilities must add up to 1…
0.10 + 0.15 + 0.15 + 0.15 + 0.20 + 0.20 + P[g] = 1
0.95 + P[g] =1, so P[g] = 0.05
(b) Probability of A is the addition of all the simple probabilities (see diagram):
P(A) =P(a)+P(b)+P(f) = 0.10+0.15+0.20 = 0.45
(c) Probability of B is the addition of all the simple probabilities (see diagram):
P(B) =P(g)+P(d)+P(e) = 0.05+0.15+0.20 = 0.40
(d) Probability of C is the addition of all the simple probabilities (see diagram):
P(C) =P(c)+P(f) = 0.15+0.20 = 0.35
(e) Probability of AB is the addition of all the simple probabilities that are in
BOTH A and B…in this case, there are NONE, so the probability is ZERO
(f) Probability of AC is the addition of all the simple probabilities that are in
BOTH A and C…in this case, there is only one P(f), so the probability is 0.20
(g) Probability of BC is the addition of all the simple probabilities that are in
BOTH B and C…in this case, there are NONE, so the probability is ZERO
(h) Probability of AB is the addition of all the simple probabilities that are in
either A or B or BOTH…in this case, that means:
P(a)+P(b)+P(f)+P(g)+P(d)+P(e) = 0.10+0.15+0.20+0.05+0.15+0.20 = 0.85
you could have said it was everything except P(c), 1 – P(c) = 1 – 0.15= 0.85
The other method for calculating this is using the Additive Law:
P(AB) = P(A) + P(B) – P(AB)
so from parts (b), (c) and (e) you get: P(AB) = 0.40 + 0.45 - 0= 0.85
Problem #98
Mutually exclusive means that the two sets do NOT Intersect (or the probability of
their intersection is ZERO). Looking at the answers to (b)-(d) you can see that A
and B are mutually exclusive and C and B are mutually exclusive (because their
intersections are both ZERO probability).
Problem #99
I would like to rewrite the probabilities in a table (which I believe will make it a bit
easier to interpret):
E1
E2
E3
E4
E5
Probability
x
x
x
x
0.40
(I have placed an “x” in the probabilities of the first four events since they are all
equal so they must all have the same value.)
Since all these probabilities must add up to 1…
x + x + x + x + 0.40 = 1
4x + 0.40 =1,
4x = 0.60, so x = 0.15 and then the table must be:
E1
Probability
0.15
(a) P[E1] = 0.15
E2
0.15
E3
0.15
E4
0.15
E5
0.40
(b) Probability of AB is the addition of all the simple probabilities that are in
either A or B or BOTH…in this case, that means:
P(E1)+P(E2)+P(E3)+P(E4)+P(E5) = 0.15+0.15+0.15+0.15+0.40 = 1.00
(c) Probability of A is the addition of all the simple probabilities (see diagram):
P(A) =P(E1)+P(E3)+P(E4) = 0.15+0.15+0.15 = 0.45
(d) Probability of B is the addition of all the simple probabilities (see diagram):
P(B) =P(E2)+P(E4)+P(E5) = 0.15+0.15+0.40 = 0.70
The Additive Law: P(AB) = P(A) + P(B) – P(AB)
so from parts (c) and (d) we have P(A) and P(B), we just need P(AB), which is the
addition of all the simple probabilities that are in BOTH A and B…in this case, the
is P(E4) = 0.15, so putting it all together you get:
P(AB) = 0.45 + 0.70 – 0.15= 1.00
Problem #100
I would like to rewrite the probabilities in a table (which I believe will make it a bit
easier to interpret):
Heads
Tails
Probability
2x
x
(I have placed an “x” in the probability of tails since it “…twice as likely to produce
a head as a tail.” And so the probability of heads must be twice that or “2x”.
Since all these probabilities must add up to 1…
2x + x = 1
3x =1, so x = 1/3 and then the table must be:
Probability
Heads
2/3 or
0.6667
Tails
1/3 or
0.3333
Problems #101-105
I would like to rewrite the probabilities in a table (which I believe will make it a bit
easier to interpret):
Probability
Poor
Average
0.14
0.37
Above
Average
0.24
Very
Good
0.18
Excellent
0.07
Problem #101
Probability of “very good or excellent” means all the probability of very good
(0.18) plus all the probability of excellent (0.07): 0.18 + 0.07 =0.25
Problem #102
Probability of “at most average” means average or less, so all the probability of
average (0.37) plus all the probability of poor (0.14): 0.37 + 0.14 =0.51
Problem #103
Probability of “at least average” means average or better, so all the probability of
average (0.37) plus all the probability of above average (0.24) plus all the
probability of very good (0.18) plus all the probability of excellent (0.07):
0.37 + 0.24 +0.18 + 0.07 =0.86
Problem #104
Probability of “neither poor nor average” means everything except poor and
average, so you subtract all the probability of average (0.37) and all the
probability of poor (0.14) from ONE (all the probability possible):
1 - 0.37 + 0.14 = 1 – 0.51 =0.49
Problem #105
Probability of “not excellent, but not poor” means everything except excellent
and poor, so you subtract all the probability of excellent (0.07) and all the
probability of poor (0.14) from ONE (all the probability possible):
1 - 0.07 + 0.14 = 1 – 0.21 =0.79
Problem #106
Since the sum of the probabilities must add to one, that is how to solve for the
“*”. 0.64+0.16+x+0.04 =1 = 0.84+x, so x = 0.16
Problem #107
“…makes at least one free throw…”, (“…at least one…” means one or more, so
either 1 or 2) so this means the “Make Miss” (1 made), “Miss Make” (1 made) and
“Make Make” (2 made) all meet the criteria.
So the probability is 0.64+0.16+0.16 = 0.96
Problem #108
“…makes at most one free throw…”, (“…at most one…” means one or less, so
either 1 or 0) so this means the “Make Miss” (1 made), “Miss Make” (1 made) and
“Miss Miss” (0 made) all meet the criteria.
So the probability is 0.04+0.16+0.16 = 0.36
Problem #109
“…makes one …”, (this means exactly one, not more, not less) so this means the
“Make Miss” (1 made) and “Miss Make” (1 made) all meet the criteria.
So the probability is 0.16+0.16 = 0.32
Problem #110
There are two possible sample sets for this question and they both work. One is a
matter of listed in order of choice, or not…so, this is QN versus QN and NQ.
Remember there is only ONE of each of the four coins, so once it is “drawn” it
cannot be drawn again.
So if I do not care about order, I get S:{PN, PD, PQ, ND, NQ, DQ}
Where if order matters, then I will get all of those above and their “reverses”:
S:{PN, PD, PQ, NP, ND, NQ, DP, DN, DQ, QP, QN, QD}
Problem #111
So count the number that have quarters (Q’s) and divide by the total number in
the sample space.
So the probability would be either: 3/6 =1/2 or 0.50 or 6/12 =1/2 or 0.50
Problem #112
“…at least 15 cents…” (“at least” meaning the least is, so 15 cents or more, in this
case) So count the number of pairs that have a value of 15 cents or more and
divide by the total number in the sample space.
S:{PN, PD, PQ, ND, NQ, DQ}
S:{PN, PD, PQ, NP, ND, NQ, DP, DN, DQ, QP, QN, QD}
So the probability would be either: 4/6 =2/3 or 0.6667 or 8/12 =2/3 or 0.6667
Problem #113
“…no more than 20 cents…” (“no more than” meaning the maximum is, so 20
cents or less, in this case) So count the number of pairs that have a value of 20
cents or less and divide by the total number in the sample space.
S:{PN, PD, PQ, ND, NQ, DQ}
S:{PN, PD, PQ, NP, ND, NQ, DP, DN, DQ, QP, QN, QD}
So the probability would be either: 3/6 =1/2 or 0.50 or 6/12 =1/2 or 0.50
Problems 114-126
In order to do these problems you need two things: (1) the sample space and
(2) the probabilities associated with each event in the sample space.
The sample space for flipping two coins is: S:{HH, HT, TH, TT} because either coin
can be heads or tails, so they both could be heads (HH) or could both be tails (TT),
or one could be heads and the other tails (HT, TH).
Now the probabilities are trickier, since they have a lot of “rules”. Let’s start by
creating a table and filling in the blanks:
HH
HT
TH
TT
Let’s start with the most complex statement: “…P[HH] = 3*P[HT]…” and let’s make
the probability of HT equal to “x”. If that is so, then by this statement, then the
probability of HH is equal to 3x…
HH
3x
HT
x
TH
TT
The next statement to look at (going backwards): “…P[HT] = P[TH]…”. So, then by
this statement, then the probability of TH is equal to x…
HH
3x
HT
x
TH
x
TT
The next statement to look at (going backwards): “Outcomes HH and TT are just
as likely as on another.” And “just as likely” means that the probabilities are
equal. So, then the probability of TT is equal to 3x…
HH
3x
HT
x
TH
x
TT
3x
Now we know that the sum of the probabilities must equal ONE, so
3x + x +x + 3x = 1 = 8x
Therefore x= 1/8, so the table looks like:
HH
HT
3/8
1/8
And now you can answer the problems…
TH
1/8
TT
3/8
Problem 114
“P[HH]” is the probability of getting both heads...which is, from above,
3/8 or 0.375
Problem 115
“P[HT]” is the probability of getting heads on the first coin and tails on the second
coin...which is, from above, 1/8 or 0.125
Problem 116
“P[A]” is the probability of all elements in set A, which is when “…one head
occurs…” or for HT or TH...which means adding those two probabilities together,
from above, 1/8 + 1/8 = 2/8 = 1/4 or 0.25
Problem 117
“P[B]” is the probability of all elements in set B, which is when “…At least one
head occurs…” or in other words, one or more heads, so one head or two heads.
So for this case it is one head (HT or TH) and also 2 heads (HH)...which means
adding these probabilities together, from above, 1/8 + 1/8 + 3/8 = 5/8 or 0.625
Problem 118
“P[C]” is the probability of all elements in set C, which is when “…Coins match…”
or in other words, both heads or both tails (HH or TT)...which means adding these
probabilities together, from above, 3/8 + 3/8 = 6/8 = 3/4 or 0.75
Problem 119
“P[D]” is the probability of all elements in set D, which is when “…At most one tail
or the coins match…”. So for this the “or” is important since that means the
union of the two statements or the combination of the two statements. It could
also be viewed as “at most one tail” or “coins match” or both happening at the
same time. The “coins match” is the answer in problem 118 (HH, TT). For “at
most one tail” that means that the maximum number of tails is one, or one or less
tails (in other words either 1 tail or zero tails)…this would result in HT and TH (for
one tail) and HH (for zero tails). So if I combine the answers I get HH, HT, TH, TT,
which is everything. And the probability of everything is 1
Problem 120
“P[AB]” is the probability of all elements in the intersection of A and B. The
intersection would be any elements that are in BOTH A and B, and since A (from
problem 116) is {HT, TH} and B (from problem 117) is {HT, TH, HH}, then the
intersection is {HT, TH}, which is the answer to problem 116: 2/8 = 1/4 or 0.25
Problem 121
“P[AC]” is the probability of all elements in the intersection of A and C. The
intersection would be any elements that are in BOTH A and C, and since A (from
problem 116) is {HT, TH} and C (from problem 118) is {TT, HH}, then there is no
intersection, so the probability is 0
Problem 122
“P[BC]” is the probability of all elements in the intersection of A and C. The
intersection would be any elements that are in BOTH A and C, and since A (from
problem 117) is {HT, TH, HH} and C (from problem 118) is {TT, HH}, then the
intersection is {HH}, so the probability(from problem 114) is 3/8 or 0.375
Problem 123
“P[AB]” is the probability of all elements in the union of A and B. The union
would be any elements that is in A or in B or in BOTH A and B, and since A (from
problem 116) is {HT, TH} and B (from problem 117) is {HT, TH, HH}, then the union
is {HT, TH, HH}, which is the answer to problem 117: 5/8 or 0.625
Problem 124
“P[AC]” is the probability of all elements in the union of A and C. The union
would be any elements that is in A or in C or in BOTH A and C, and since A (from
problem 116) is {HT, TH} and C (from problem 118) is {TT, HH}, then the union is
{TT, HT, TH, HH}, which is all the elements, so the probability is: 1
Problem 125
“P[BC]” is the probability of all elements in the union of B and C. The union
would be any elements that is in B or in C or in BOTH B and C, and since B (from
problem 117) is {HT, TH, HH} and C (from problem 118) is {TT, HH}, then the union
is {TT, HT, TH, HH}, which is all the elements, so the probability is: 1
Problem 126
“…mutually exclusive…” means that the two sets exclude one another, or do not
share any elements, or as a quick way of knowing, there is no intersection
between the two sets. The possible pairs are A-B, A-C, A-D, B-C, B-D, and C-D. A
and B have an intersection (problem 120), but A and C do not have an intersection
(problem 121) so A and C are mutually exclusive. Since D contains everything,
anything with D will have an intersection, leaving only B and C and it has an
intersection (problem 122). So, the answer is that only A and C are mutually
exclusive.
Problem 127
The key word in setting the sample space is the work “rank” or in other words,
order matters…it does matter which one is first versus which one is last. So this
will be the same as the answer for problem 67 (for all the same reasons), just with
different letters: S: {BSN, BNS, SBN, SNB, NSB, NBS}
Now with this information you can answer problems 128-130…
Problem 128
“...ranks B first…” gives us all elements with B up front (BSN, BNS), so there are 2
possible answers; and to get the probability you divide by the total number of
outcomes (6): 2/6 = 1/3 or 0.3333
Problem 129
“...ranks S last…” gives us all elements with S at the end (BNS, NBS), so there are 2
possible answers; and to get the probability you divide by the total number of
outcomes (6): 2/6 = 1/3 or 0.3333
Problem 130
“...ranks N first and S not last…” is an intersection of the two statements because
of the word “and”, so BOTH things must occur in the same element. So of the
elements that have N first (NSB, NBS), there is only one (NSB) in which S is not
last. So with the one to get the probability you divide it by the total number of
outcomes (6): 1/6 or 0.1667
Problem 131
Since we are talking about admission or not for college A, and the same for
college B…iy should bring to mind a 20-circle Venn Diagram… Since there are only
two events, so it is a two circle Venn Diagram:
A
B
Next, let’s see what it says, “…admission to college A is 0.8…” (so both parts of
circle A must add up to be 0.80).
0.80
A
A
B
B
Next, it says “…admission to college B is 0.5…” which means that the intersection
of A and B is 0.1…
A
A
0.80
B
0.50
The last part is the “tricky” part; it says “…rejection by at least one of the two
colleges is 0.6…” Up to this point we have talked about “admission” not
“rejection”, so to make this easier, what is the statement if was talking about
“admission”? Well, first it is talking about “at least one”, which means one or
more rejections…so one rejection or two rejections. Well if rejected by two
colleges (out of two colleges) means admission to none (zero admissions); and for
one rejection, that means also one admission…so the only thing left is 2
admissions…so that becomes the opposite (or complement) and the complement
is 1 – 0.6 = 0.4. So then the admission to both colleges (two colleges) is 0.4.
0.80
A
B
M
N
0.4
K
B
A
0.50
Now the rest of the segments can be solved for…if all of A is 0.8, but the
intersection is 0.4, then the other part must also be 0.4to add up to that. And
likewise for B, the other segment must be 0.5-0.4 = 0.1. And finally, all the parts
must add up to 1 so the outside must be 1 – 0.4 – 0.4 – 0.1 = 0.1…which gives us
the following Venn Diagram from which to answer the question:
0.80
A
B
M
N
0.4
0.4 0.1
K
B
A
0.50
0.1
The question is “What is the probability that he will be admitted by at least one of
the two colleges?” “…at least one…” means one or more so accepted for
admission in either one college or two colleges, so it is all the probability inside
either of the circle (it is the union) so the answer is: 0.4 +0.4 +0.1 = 0.9
Problem 132
The Additive Law for 2 sets (A and B) is P[AB] = P[A] + P[B] – P[AB]. So it starts
with adding all the “circles” and then subtracting the intersection”; so we will
start with that same philosophy for the Additive Law for 3 sets (A, B and C):
P[ABC] = P[A] + P[B] + P[C] – P[AB] – P[AC] – P[BC], (note that with three
circles there are 3 intersections), but to use the letters, I will eliminate the
probability part, giving an equation of: ABC = A + B + C – AB – AC – BC
Looking at the diagram, you can see that the union of all three circles [ABC] is
all the elements in those circles, which is: {a, b, d, e, c, f, g}. Also, from the
diagram, A = {a, b, d, e}, B = {b, c, e, f}, C = {d, e, f, g}, AB = {b, e}, AC = {d, e},
and BC = {e, f}…please note that the intersection of any 2 circles results in a
“football” shaped section which has two parts; and that is why the intersections
have 2 letters in them.
Now we will do the adding and subtracting and see if the equation is correct:
a+b+d+e + b+c+e+f + d+e+f+g – b-e – d-e – e-f = a, b, d, c, f, g
Which is not quite right, we are short the “e” or the intersection of all three
circles (it seems to have been eliminated when we subtracted all the
intersections).
So the correct equation is:
P[ABC] = P[A] + P[B] + P[C] – P[AB] – P[AC] – P[BC] + P[ABC]
Problems 133-134
“The Odds” can be simply defined as how many for and how many against a
certain outcome. Let’s do an example. Say 100 students must vote for
something…they must vote either “yes” or “no”. Now let’s say that 55 of these
students vote “yes” (which means that 45 voted “no”). So based on this data the
odds of the vote being “yes” is 55:45 (or you could simplify it to 11:9). But what if
you wanted to know the probability? Well, remember that probability is how
many answer the question (in this example, voted “yes”…55) divided by the total
(100): 55/100 = 11/20 or 0.55
Problem 133
So probability of selling is 0.85 (per the text)…which is the same as 85/100. Since
probability is how many to sell divided by the total, then 85 out of every 100 will
buy a ticket (or be sold to), so 15 would not buy. So the odds are: 85:15
Problem 134
If the odds for selling out is 85:15 (from problem 133), then for NOT selling out is
the opposite…or 15:85
Problem 135
Remember that probability is how many answer the question (in this case Boris
wins, 175) divide by the total (184): 175/184 or 0.9511
Problem 136
For winning 175 games there was also 9 losses in the 184 games, so the odds of
winning are: 175:9
Problem 137
Your probability of passing is 0.94 or 94/100…that means that there is a 6/100
chance of you not passing. In other words out of 100 tries, 94 times you would
pass and 6 tries you would fail, so the odds are: 6:94 (remember you were asked
for the odds of failing, so that number goes first)