Working with Probabilities Physics 115a (Slideshow 1) A. Albrecht These slides related to Griffiths...
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Working with Probabilities Physics 115a (Slideshow 1) A. Albrecht These slides related to Griffiths section 1.3
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Transcript of Working with Probabilities Physics 115a (Slideshow 1) A. Albrecht These slides related to Griffiths...
Working with ProbabilitiesPhysics 115a (Slideshow 1)
A. Albrecht
These slides related to Griffiths section 1.3
Consider the following group of people in a room:
Age Number
14 1
15 1
16 3
22 2
24 2
25 5
Histogram Form
0 10 20 30 400
1
2
3
4
5
Age
Nu
mb
er
Consider the following group of people in a room:
Age Number
14 1
15 1
16 3
22 2
24 2
25 5
Total people = 14
Consider the following group of people in a room:
Total people = 14
Age Number Probability
14 1
15 1 ?
16 3
22 2
24 2
25 5
Consider the following group of people in a room:
Total people = 14
Age Number Probability
14 1
15 1 1/14
16 3
22 2
24 2
25 5
Consider the following group of people in a room:
Total people = 14
Age Number Probability
14 1
15 1 1/14
16 3
22 2
24 2
25 5 ?
Consider the following group of people in a room:
Total people = 14
Age Number Probability
14 1
15 1 1/14
16 3
22 2
24 2
25 5 5/14
Consider the following group of people in a room:
Total people = 14
Age Number Probability
14 1 ?
15 1 1/14
16 3 ?
22 2 ?
24 2 ?
25 5 5/14
Consider the following group of people in a room:
Total people = 14
Age Number Probability
14 1 1/14
15 1 1/14
16 3 3/14
22 2 2/14
24 2 2/14
25 5 5/14
Probability Histogram
0 10 20 30 400
0.1
0.2
0.3
0.4
Age
Pro
ba
bili
ty
0 10 20 30 400
1
2
3
4
5
Age
Nu
mb
er
Number Histogram
NB: The probabilities for ages not listed are all zero
Total people = 14
Age Number Probability
14 1 1/14
15 1 1/14
16 3 3/14
22 2 2/14
24 2 2/14
25 5 5/14
Assuming Age<20, what is the probability of finding each age?
Total people = 14
Age Number Probability
14 1 ?
15 1 ?
16 3 ?
22 2 ?
24 2 ?
25 5 ?
Assuming Age<20, what is the probability of finding each age?
Total people = 14
Age Number Probability
14 1 ?
15 1 ?
16 3 ?
22 2 0
24 2 0
25 5 0
Assuming Age<20, what is the probability of finding each age?
Total people = 14
Age Number Probability
14 1 1/5
15 1 1/5
16 3 3/5
22 2 0
24 2 0
25 5 0
Total people = 14
Age Number Probability
14 1 1/14
15 1 1/14
16 3 3/14
22 2 2/14
24 2 2/14
25 5 5/14
Assuming no age constraint, what is the probability of finding each age?
Related to collapse of the waveunction (“changing the question”)
Assuming Age<20, what is the probability of finding each age?
Total people = 14
Age Number Probability
14 1 1/5
15 1 1/5
16 3 3/5
22 2 0
24 2 0
25 5 0
Related to collapse of the waveunction (“changing the question”)
Consider a different room with different people:
Age Number
19 3
20 2
21 5
22 3
24 1
25 1
Total people = 15
Consider a different room with different people:
Age Number Probability
19 3 3/15
20 2 2/15
21 5 5/15
22 3 3/15
24 1 1/15
25 1 1/15
Total people = 15
0 10 20 30 400
1
2
3
4
5
Age
Nu
mb
er
Red Room Numbers
Red Room Probabilities
0 10 20 30 400
0.1
0.2
0.3
Age
Pro
ba
bili
ty
Combine Red and Blue rooms
Total people = 29
Age Number Probability
14 1 1/29
15 1 1/29
16 3 3/29
19 3 3/29
20 2 2/29
21 5 5/29
22 2+3 5/29
24 2+1 3/29
25 5+1 6/29
0 10 20 30 400
1
2
3
4
5
6
Age
Nu
mb
er
0 10 20 30 400
0.1
0.2
Age
Pro
ba
bili
ty
0 10 20 30 400
1
2
3
4
5
6
Age
Nu
mb
er
0 10 20 30 400
0.1
0.2
Age
Pro
ba
bili
ty
Lessons so far
• A simple application of probabilities
• Normalization
• “Re-Normalization” to answer a different question
• Adding two “systems”.
• All of the above are straightforward applications of intuition.
Expectation Values
0 10 20 30 400
0.1
0.2
0.3
0.4
Age
Pro
ba
bili
ty
Most probable answer = 25Median = 23
Average = 21
0 10 20 30 400
0.1
0.2
0.3
0.4
Age
Pro
ba
bili
ty
Most probable answer = 25Median = 23
Average = 21
Lesson: Lots of different types of questions (some quite similar) with different answers. Details depend on the full probability distribution.
Average (mean):
0
0
j
jtot
jN j
j jP jN
• Standard QM notation
• Called “expectation value”
• NB in general (including the above) the “expectation value” need not even be possible outcome.
Average (number squared)
Age Number (Number)2 Probability
14 1 1 1/14
15 1 1 1/14
16 3 9 3/14
22 2 4 2/14
24 2 4 2/14
25 5 25 5/14
2
02 2
0
449.4j
jtot
j N j
j j P jN
In general, the average (or expectation value) of some function f(j) is
0j
f j f j P j
Careful: In general 2 2j j
441 449.4
The “width” of a probability distribution
1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
Age
Nu
mb
er
1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
Age
Nu
mb
er
1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
Age
Nu
mb
er
1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
Age
Nu
mb
er
Discuss eqns 1.10 through 1.13 at board
Continuous Variables
Continuous Variables
Why not measure age in weeks?
0 500 1000 15000
0.2
0.4
0.6
0.8
1
Age (Weeks)
Nu
mb
er
Blue room in weeks
0 500 1000 15000
0.02
0.04
0.06
0.08
Age (Weeks)
Pro
ba
bili
tyBlue room in weeks
Conclusion: Blue room in weeks not very useful/intuitive
Another case where a measure of age in weeks might by useful:
The ages of students taking health in the 8th grade in a large school district (3000 students).
0 200 400 600 800 10000
5
10
15
20
25
Age (Weeks)
Nu
mb
er
0 200 400 600 800 10000
2
4
6
8
x 10-3
Age (Weeks)
Pro
ba
bili
ty
0 200 400 600 800 10000
2
4
6
8
x 10-3
Age (Weeks)
Pro
ba
bili
ty
0 200 400 600 800 10000
0.1
0.2
0.3
0.4
Age (Weeks)
Pro
ba
bili
ty p
er
yea
r
