TEST BANK
MOLECULAR DRIVING FORCES
1
,Chapter 1
Principles Of Probability
1. Combining Independent Probabilities.
(a) The Simplest Way To Solve This Problem Is To Recall That When Probabilities Are
Independent, And You Want The Probability Of Events A AND B, You Can Multiply
Them. When Events Are Mutually Exclusive And You Want The Probability Of Events
A OR B, You Can Add The Probabilities. Therefore We Try To Structure The Problem
Into An AND And OR Problem. We Want The Probability Of Getting Into H OR DSM
Or UCSF. But This Doesn’t Help Because These Events Are Not Mutually Exclusive
(Mutually Exclusive Means That If One Happens, The Other Cannot Happen). So We
Try Again. The Probability Of Acceptance Somewhere, P (A), Is P (A) = 1 − P (R), Where
P (R) Is The Probability That You’re Rejected Everywhere. (You’re Either Accepted
Somewhere Or You’re Not.) But This Probability Can Be Put In The Above Terms. P (R)
= The Probability That You’re Rejected At H AND At DSM AND At UCSF. These Events
Are Independent, So We Have The Answer. The Probability Of Rejection At H Is
P(Rh) = 1 − 0.5 = 0.5. Rejection At DSM Is P(Rdsm) = 1 − 0.3 = 0.7. Rejection At
UCSF Is P(Rucsf) = 1 − 0.1 = 0.9. Therefore P (R) = (0.5)(0.7)(0.9) = 0.315. Therefore
The Probability Of At Least One Acceptance
= P (A) = 1 − P (R) = 0.685.
2
,(b) The Simple Answer Is That This Is The Intersection Of Two Independent Events:
P(Ah)P(Adsm) = (0.50)(0.30)
= 0.15.
A More Mechanical Approach To Either Part (A) Or This Part Is To Write Out All The
Possible Circumstances. Rejection And Acceptance At H Are Mutually Exclusive. Their
Probabilities Add To One. The Same For The Other Two Schools. Therefore All Possible
Circumstances Are Taken Into Account By Adding The Mutually Exclusive Events
Together, And Multiplying Independent Events:
[P(Ah) + P(Rh)][P(Adsm) + P(Rdsm)][P(Aucsf) + P(Rucsf)] = 1,
Or Equivalently,
= P(Ah)P(Adsm)P(Aucsf) + P(Ah)P(Adsm)P(Rucsf)
+P(Ah)P(Rdsm)P(Aucsf) + · · ·
Where The First Term Is The Probability Of Acceptance At All 3, The Second Term
Represents Acceptance At H And DSM But Rejection At UCSF, The Third Term
Represents Acceptance At H And UCSF But Rejection At DSM, Etc. Each Of These
Events Is Mutually Exclusive With Respect To Each Other; Therefore They Are All
Added. Each Individual Term Represents Independent Events Of, For Example, Ah And
Adsm And Aucsf. Therefore It Is Simple To Read Off The Answer In This Problem: We
Want Ah And Adsm, But Notice We Don’t Care About UCSF. This Probability Is
P(Ah)P(Adsm) = P(Ah)P(Adsm)[P(Aucsf) + P(Rucsf)]
= (0.50)(0.30)
= 0.15.
Note That We Could Have Solved Part (A) The Same Way; It Would Have Required
Adding Up All The Appropriate Possible Mutually Exclusive Events. You Can Check That
It Gives The Same Answer As Above (But Notice How Much More Tedious It Is).
3
, 2. Probabilities Of Sequences.
of nine monomers.
(a) Each Base Occurs With Probability 1/4. The Probability Of An A In Position 1 Is 1/4,
Of A In Position 2 Is 1/4, Of A In Position 3 Is 1/4, Of F In Position 4 Is 1/4, And So
On. There Are 9 Bases. The Probability Of This Specific Sequence Is (1/4)9 = 3.8 ×
10−6.
(b) Same Answer As (A) Above.
(c) Each Specific Sequence Has The Probability Given Above, But In This Case There Are
Many Possible Sequences Which Satisfy The Requirement That We Have 4 A’s, 2 F ’S, 2
J’s, And 1
C. How Many Are There? We Start As We Have Done Before, By Assuming All Nine
Objects Are Distinguishable. There Are 9! Arrangements Of Nine Distinguishable
Objects In A Linear Sequence. (The First One Can Be In Any Of Nine Places, The Second
In Any Of The Remaining Eight Places, And So On.) But We Can’t Distinguish The Four
A’s, So We Have Overcounted By A Factor Of 4!, And Must Divide This Out. We Can’t
Distinguish The Two F ’S, So We Have Overcounted By 2!, And Must Also Divide This
Out. And So On. So The Probability Of Having This Composition Is
" # 9
9! 1
0.014.
4
4
MOLECULAR DRIVING FORCES
1
,Chapter 1
Principles Of Probability
1. Combining Independent Probabilities.
(a) The Simplest Way To Solve This Problem Is To Recall That When Probabilities Are
Independent, And You Want The Probability Of Events A AND B, You Can Multiply
Them. When Events Are Mutually Exclusive And You Want The Probability Of Events
A OR B, You Can Add The Probabilities. Therefore We Try To Structure The Problem
Into An AND And OR Problem. We Want The Probability Of Getting Into H OR DSM
Or UCSF. But This Doesn’t Help Because These Events Are Not Mutually Exclusive
(Mutually Exclusive Means That If One Happens, The Other Cannot Happen). So We
Try Again. The Probability Of Acceptance Somewhere, P (A), Is P (A) = 1 − P (R), Where
P (R) Is The Probability That You’re Rejected Everywhere. (You’re Either Accepted
Somewhere Or You’re Not.) But This Probability Can Be Put In The Above Terms. P (R)
= The Probability That You’re Rejected At H AND At DSM AND At UCSF. These Events
Are Independent, So We Have The Answer. The Probability Of Rejection At H Is
P(Rh) = 1 − 0.5 = 0.5. Rejection At DSM Is P(Rdsm) = 1 − 0.3 = 0.7. Rejection At
UCSF Is P(Rucsf) = 1 − 0.1 = 0.9. Therefore P (R) = (0.5)(0.7)(0.9) = 0.315. Therefore
The Probability Of At Least One Acceptance
= P (A) = 1 − P (R) = 0.685.
2
,(b) The Simple Answer Is That This Is The Intersection Of Two Independent Events:
P(Ah)P(Adsm) = (0.50)(0.30)
= 0.15.
A More Mechanical Approach To Either Part (A) Or This Part Is To Write Out All The
Possible Circumstances. Rejection And Acceptance At H Are Mutually Exclusive. Their
Probabilities Add To One. The Same For The Other Two Schools. Therefore All Possible
Circumstances Are Taken Into Account By Adding The Mutually Exclusive Events
Together, And Multiplying Independent Events:
[P(Ah) + P(Rh)][P(Adsm) + P(Rdsm)][P(Aucsf) + P(Rucsf)] = 1,
Or Equivalently,
= P(Ah)P(Adsm)P(Aucsf) + P(Ah)P(Adsm)P(Rucsf)
+P(Ah)P(Rdsm)P(Aucsf) + · · ·
Where The First Term Is The Probability Of Acceptance At All 3, The Second Term
Represents Acceptance At H And DSM But Rejection At UCSF, The Third Term
Represents Acceptance At H And UCSF But Rejection At DSM, Etc. Each Of These
Events Is Mutually Exclusive With Respect To Each Other; Therefore They Are All
Added. Each Individual Term Represents Independent Events Of, For Example, Ah And
Adsm And Aucsf. Therefore It Is Simple To Read Off The Answer In This Problem: We
Want Ah And Adsm, But Notice We Don’t Care About UCSF. This Probability Is
P(Ah)P(Adsm) = P(Ah)P(Adsm)[P(Aucsf) + P(Rucsf)]
= (0.50)(0.30)
= 0.15.
Note That We Could Have Solved Part (A) The Same Way; It Would Have Required
Adding Up All The Appropriate Possible Mutually Exclusive Events. You Can Check That
It Gives The Same Answer As Above (But Notice How Much More Tedious It Is).
3
, 2. Probabilities Of Sequences.
of nine monomers.
(a) Each Base Occurs With Probability 1/4. The Probability Of An A In Position 1 Is 1/4,
Of A In Position 2 Is 1/4, Of A In Position 3 Is 1/4, Of F In Position 4 Is 1/4, And So
On. There Are 9 Bases. The Probability Of This Specific Sequence Is (1/4)9 = 3.8 ×
10−6.
(b) Same Answer As (A) Above.
(c) Each Specific Sequence Has The Probability Given Above, But In This Case There Are
Many Possible Sequences Which Satisfy The Requirement That We Have 4 A’s, 2 F ’S, 2
J’s, And 1
C. How Many Are There? We Start As We Have Done Before, By Assuming All Nine
Objects Are Distinguishable. There Are 9! Arrangements Of Nine Distinguishable
Objects In A Linear Sequence. (The First One Can Be In Any Of Nine Places, The Second
In Any Of The Remaining Eight Places, And So On.) But We Can’t Distinguish The Four
A’s, So We Have Overcounted By A Factor Of 4!, And Must Divide This Out. We Can’t
Distinguish The Two F ’S, So We Have Overcounted By 2!, And Must Also Divide This
Out. And So On. So The Probability Of Having This Composition Is
" # 9
9! 1
0.014.
4
4
