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.

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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%
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%
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