Example: Consider the case in which the probability of success in a single trial is 4 and four trials are to be made. Evaluate the individual and cumulative probabilities of success in this case.
Example: Consider the case in which the probability of success in a single trial is 4 and four trials are to be made. Evaluate the individual and cumulative probabilities of success in this case.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
Related questions
Topic Video
Question
100%
help me to uderstan this question?

Transcribed Image Text:Part-6-2 - PowerPoint
mo abd
File
Home
Insert
Design
Transitions
Animations
Slide Show
Review
View
Recording
Help
Tell me what you want to do
& Share
| Layout v
A A E: -
O Shape Fill
P Find
AZL
L/ Shape Outline
ab. Replace
Reset
Arrange Quick
Styles
Paste
New
BIUS abe AV - Aa - aby - A
Slide
A Section v
O Shape Effects
h Select -
Clipboard
Slides
Font
Paragraph
Drawing
Editing
4
Binomial Distribution
Binomial Distribution
Example: Consider the case in which the probability of success in a single trial is 4
and four trials are to be made. Evaluate the individual and cumulative probabilities of
success in this case.
Example: Consider the case in which the probability of success in a single trial is ¼
and four trials are to be made. Evaluate the individual and cumulative probabilities of
n= 4,
p=0.25,
(p+ q)* = pt + 4p°q+ 6p°q² + 4pq
q=1-0.25 = 0.75
(P+9)%)
success in this case.
1
Number of
Cumulative
n = 4,
p = 0.25,
q=1– 0.25 = 0.75
Individual prob ability
= (3/4)*
4pg = 4(1/4)(3/4) +108/256
6p'q = 6(1/4) (3/4)
4p'q = 4(1/4) (3/4)
= (14)
(P+9)%)
probability
81/256
189/256
243/256
successes failures
4
81/256
(p + q)* = p4 +4p°q+ 6p?q² + 4pd° + q*
3
54/256
12/256
1/256
E=1
255/256
256/256
4
1
Number of
Cumulative
Saturday, 31 July 2021
Reliability Evaluation of Engineering Systemis
Individual probability
q*
failures
probability
successes
= (3/4)
4pg', - 4(1/4)(3/4)
6p'q = 6(1/4) (3/4)*- 54/256
4p'q = 4(1/4) (3/4)
=(1/4)
2
4
81/256
81/256
Probability Distributions
1
108/256
189/256
2
243/256
Probability Density Function
Probability Di stribution Function
3
1
12/256
255/256
1.0
0.4
0.8
1/256
256/256
0.3
0.8
E=1
0.2
0.4
0.
0.2
0.0
0.0
1 2
# of Heads, x
1
2 3
3
4
# of Heads, x
Saturday, 31 July 2021
Notes
Reliability Evaluation of Engineering Systemis
Slide 4 of 12
Arabic (Iraq)
= Notes
Comments
+ 66%
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 1 images

Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below.Recommended textbooks for you

A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON


A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
