The document provides solutions to probability questions involving throwing ninja stars.
For the first question, the probability of at least two stars being black if three stars are thrown without replacement is calculated to be 40%.
The second question calculates the probability of the third star being red as 25%.
The third question considers a scenario where the colors are replaced after each throw, and calculates the probability of the second star being blue as 32%.
Imagine a rectangle on a dot paper. Suppose it is a pool table.
Investigate the path of a ball which starts at one corner of the table, is pushed to an edge, bounces off that edge to another, and so on, as shown in the diagram. When the ball finally reaches a corner it drops off the table.
In this investigation, some aspects were examined:
a. the account of the aspect involving the numbers of dots in a column or/and in a row;
b. the figures drawn for the cases considered;
c. the table showing the data obtained from the investigation;
d. the patterns observed
e. the presentation of conjectures from A to M;
f. the testing of conjectures from A to M, and;
g. the elaboration of this investigation.
Imagine a rectangle on a dot paper. Suppose it is a pool table.
Investigate the path of a ball which starts at one corner of the table, is pushed to an edge, bounces off that edge to another, and so on, as shown in the diagram. When the ball finally reaches a corner it drops off the table.
In this investigation, some aspects were examined:
a. the account of the aspect involving the numbers of dots in a column or/and in a row;
b. the figures drawn for the cases considered;
c. the table showing the data obtained from the investigation;
d. the patterns observed
e. the presentation of conjectures from A to M;
f. the testing of conjectures from A to M, and;
g. the elaboration of this investigation.
JĒZUS KRISTUS KOMANDU:
KRISTIEŠI! PAMEST! PAMEST KRIEVIJU! BĒGT, PUTINS! PAMEST VALSTI PSRS! PAMEST VALSTI ATEISTI! BĒGT, IZRAĒLA! PAMEST LIBĀNU! MESTOS VALSTĪS VISĀ IZRAĒLĀ! PASAULES KARA! NUCLEAR KARA! KODOLENERĢIJAS MĀKONIS! SĪRIJU ! JERUSALEM ! GAZA ! AMERIKĀŅI ! IRĀNA ! ISLĀMA VALSTI ! ISLĀMU ! CHRISTIAN ! ANTIKRISTS ! ELIJA ! PRAVIEŠI ! YHWH ! JĒZUS KRISTUS IR AIZLIEGTS ĒST DZĪVNIEKUS
What Nobody's Telling You About Agile and DevOpsTasktop
Everyone is talking about improving software delivery using Agile and DevOps. They've had some success - but the secret nobody is talking about is that it's not really working at enterprise scale.
In this talk, we discuss:
* the common goals of Agile and DevOps transformations
* how these goals break down at enterprise scale
* how you can achieve an integrated value stream that will put your transformation back on track.
This presentation uses the technology of Microsoft Multiple Mouse Mischief software. If you need assistance, visit microsoft site for multiple mouse support. Probability aptitude questions level 2
JĒZUS KRISTUS KOMANDU:
KRISTIEŠI! PAMEST! PAMEST KRIEVIJU! BĒGT, PUTINS! PAMEST VALSTI PSRS! PAMEST VALSTI ATEISTI! BĒGT, IZRAĒLA! PAMEST LIBĀNU! MESTOS VALSTĪS VISĀ IZRAĒLĀ! PASAULES KARA! NUCLEAR KARA! KODOLENERĢIJAS MĀKONIS! SĪRIJU ! JERUSALEM ! GAZA ! AMERIKĀŅI ! IRĀNA ! ISLĀMA VALSTI ! ISLĀMU ! CHRISTIAN ! ANTIKRISTS ! ELIJA ! PRAVIEŠI ! YHWH ! JĒZUS KRISTUS IR AIZLIEGTS ĒST DZĪVNIEKUS
What Nobody's Telling You About Agile and DevOpsTasktop
Everyone is talking about improving software delivery using Agile and DevOps. They've had some success - but the secret nobody is talking about is that it's not really working at enterprise scale.
In this talk, we discuss:
* the common goals of Agile and DevOps transformations
* how these goals break down at enterprise scale
* how you can achieve an integrated value stream that will put your transformation back on track.
This presentation uses the technology of Microsoft Multiple Mouse Mischief software. If you need assistance, visit microsoft site for multiple mouse support. Probability aptitude questions level 2
JEE Mathematics / Lakshmikanta Satapathy / Questions and Answers on Probability involving Independent events, mutually exclusive events in experiments of drawing colored balls from bags and problem solving probability of students using Addition theorem and Multiplication theorem
Each of the six equally spaced points on this circle has
been joined to a point that is two points away from
it in a clockwise direction. The result can be called
a [6,2] figure.
Equally spaced points on a circle can be joined by chords in various ways.
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Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
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The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
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https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
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Welcome to ViralQR, your best QR code generator available on the market!
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Our Vision
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Static QR Codes: Create free static QR codes. These QR codes are able to store significant information such as URLs, vCards, plain text, emails and SMS, Wi-Fi credentials, and Bitcoin addresses.
Dynamic QR codes: These also have all the advanced features but are subscription-based. They can directly link to PDF files, images, micro-landing pages, social accounts, review forms, business pages, and applications. In addition, they can be branded with CTAs, frames, patterns, colors, and logos to enhance your branding.
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Additionally, there is a 14-day free offer to ViralQR, which is an exceptional opportunity for new users to take a feel of this platform. One can easily subscribe from there and experience the full dynamic of using QR codes. The subscription plans are not only meant for business; they are priced very flexibly so that literally every business could afford to benefit from our service.
Why choose us?
ViralQR will provide services for marketing, advertising, catering, retail, and the like. The QR codes can be posted on fliers, packaging, merchandise, and banners, as well as to substitute for cash and cards in a restaurant or coffee shop. With QR codes integrated into your business, improve customer engagement and streamline operations.
Comprehensive Analytics
Subscribers of ViralQR receive detailed analytics and tracking tools in light of having a view of the core values of QR code performance. Our analytics dashboard shows aggregate views and unique views, as well as detailed information about each impression, including time, device, browser, and estimated location by city and country.
So, thank you for choosing ViralQR; we have an offer of nothing but the best in terms of QR code services to meet business diversity!
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Monitoring and observability aren’t traditionally found in software curriculums and many of us cobble this knowledge together from whatever vendor or ecosystem we were first introduced to and whatever is a part of your current company’s observability stack.
While the dev and ops silo continues to crumble….many organizations still relegate monitoring & observability as the purview of ops, infra and SRE teams. This is a mistake - achieving a highly observable system requires collaboration up and down the stack.
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https://alandix.com/academic/papers/synergy2024-epistemic/
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Sr Director, Infrastructure Ecosystem, Arm.
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2. Part I: Shinobi has 16 ninja stars. Four (4) of them are red, five (5) of them are blue and seven (7) of them are black. If Shinobi had thrown 3 stars without replacing them, what would be the probability of the following? a) What is the probability of at least 2 stars being black? b) What is the probability of the third star being red?
3. Part I: a) What is the probability of at least 2 stars beeing black? The colour of a succeeding star depends on the colour of the star previous to it. We call this dependent probability where the outcome of the second step depends on the outcome of the first step . This is oppose to independent probability where the outcome of the second step does not depend on the outcome of the first . The most efficient way to look at a problem like this visually is to draw a tree diagram.On the tree diagram, you would draw out all the possiblities in each step and then insert the probability for each outcome as shown in the next slide.
4. R Bl B R Bl B R Bl B R Bl B B 5 16 R Bl B R Bl B R Bl B R Bl B R 4 16 3 15 7 15 R Bl B R Bl B R Bl B R Bl B Bl 5 15 2 14 7 14 5 14 3 14 6 14 7 14 3 14 15 4 7 15 4 15 3 14 7 14 6 14 7 14 4 15 4 14 4 14 4 14 4 14 4 14 3 14 6 15 5 15 3 14 7 16 5 14 5 14 4 14 5 14 4 14 6 14 6 14 4 14 R = Red (4) B = Blue (5) Bl = Black (7)
5. Please note that where the Red, Blue, and Black start branching off individually they themselves are suppose to have branched off from the same initial spot like this: However, we did not have room on the slide to illustrate this at its best size. For this we apologize, but we hope it does not confuse you too much. R B Bl
6. R Bl B R Bl B R Bl B R Bl B B Look for the sequences where there are two (2) black stars. 5 16 R Bl B R Bl B R Bl B R Bl B R 4 16 3 15 7 15 R Bl B R Bl B R Bl B R Bl B Bl 5 15 2 14 7 14 5 14 3 14 6 14 7 14 3 14 15 4 7 15 4 15 3 14 7 14 6 14 7 14 4 15 4 14 4 14 4 14 4 14 4 14 3 14 6 15 5 15 3 14 7 16 5 14 5 14 4 14 5 14 4 14 6 14 6 14 4 14 5 14
7. For the probability of 2 black stars in a sequence look for the sequence of branches with at least 2 blacks. -Since it starts of being out of 16 stars, the denominator of the first branch is 16. As the stars aren't replaced, the denominator goes down one unit because of the star we took. -As you move down each branch you can calculate the probability of each sequence. -Moving down the branch, you take the probability of each outcome in a sequence and multiply it by each other. -You the take that total and add it to the total of the other sequences that have 2 black
9. = = 0.4% = 40% There is a 40% probability that when three stars are thrown, at least two stars are black. 2 5 = + ( ) 210 3360 168 3360 ( ) + ( ) 210 3360 + 168 3360 ( ) ( ) 210 3360 + ( ) 210 3360 + + 168 3360 ( ) 1344 3360
10. Shinobi has 16 ninja stars. Four (4) of them are red, five (5) of them are blue and seven (7) of them are black. If Shinobi had thrown 3 stars without replacing them, what would be the probability of the following? b) What is the probability of the third star being red?
11. This is again, dependent probability because the outcome of a star depends on the star preceeding it. We will be using another tree diagram to help you visualize the problem. We are looking for the probability of the third star being red. In total, the number of stars we have is 16. That's 4 red, 5 blue, and 7 black. Remember that as each star is taken, it is not replace.
12. R Bl B R Bl B R Bl B R Bl B B 5 16 R Bl B R Bl B R Bl B R Bl B R 4 16 3 15 7 15 R Bl B R Bl B R Bl B R Bl B Bl 2 14 7 14 5 14 3 14 6 14 7 14 3 14 15 4 7 15 4 15 3 14 7 14 6 14 7 14 4 15 4 14 4 14 4 14 4 14 4 14 3 14 6 15 5 15 3 14 7 16 5 14 5 14 4 14 5 14 4 14 6 14 6 14 4 14 5 14 Note: Look for the the sequences where the red is at the last branch. 5 15
13. When you are looking for the probability of the third star being red, you look for the sequences where the last branch has a red. -Since it starts of being out of 16 stars, the denominator of the first branch is 16. As the stars aren't replaced, the denominator goes down one unit because of the star we took. -As you move down each branch you can calculate the probability of each sequence. -Moving down the branch, you take the probability of each outcome in a sequence and multiply it by each other. -You the take that total and add it to the total of the other sequences that have red for the last star.
14. Calculating the probability of the third star being red... P ( R , R , R ) P ( R , B , R ) P ( R , Bl, R ) P ( B , R , R ) P ( B , B , R ) P ( B , Bl, R ) P (Bl, Bl, R ) P (Bl, B , R ) P (Bl, R , R ) = = = = = = = = = = = = = = = = = = 24 3360 ( ) 16 15 14 4 3 2 ( ) ( ) ( ) 4 5 3 16 15 14 ( ) ( ) ( ) 4 7 3 16 15 14 ( ) ( ) ( ) 5 4 3 16 15 14 ( ) ( ) ( ) 5 4 4 16 15 14 ( ) ( ) ( ) 5 7 4 16 15 14 ( ) ( ) ( ) 7 6 4 16 15 14 ( ) ( ) ( ) 7 5 4 16 15 14 ( ) ( ) ( ) 7 4 3 16 15 14 ( ) ( ) ( ) 60 3360 ( ) 84 3360 ( ) 60 3360 ( ) 80 3360 ( ) 140 3360 ( ) 168 3360 ( ) 140 3360 ( ) 84 3360 ( )
15. = = 0.25 = 25% 1 4 = 3360 840 24 3360 ( ) 60 3360 ( ) 84 3360 ( ) 60 3360 ( ) 80 3360 ( ) 140 3360 ( ) 168 3360 ( ) 140 3360 ( ) 84 3360 ( ) + + + + + + + + There is a 25% probability that the third star is red.
16. Part II: Shinobi has 16 ninja stars. Four (4) of them are red, five (5) of them are blue and seven (7) of them are black. 2 stars are thrown. If a blue star thrown, it is discarded and replaced with a red star. If a red star is thrown, it is discarded and replaced with a black star. If a black star is thrown, it is discarded and replaced with a blue star. What is the probability of the second star being thrown is blue?
17. What we know from the problem... This is a dependent probability because the outcome of the second star depends on the outcome of the second star. We also know that when a star is thrown: Blue is replaced with Red . Red is replaced with Black . Black is replaced with Blue . We will start off by drawing another tree diagram and insert the probability of each outcome. Remember, for this question we are looking for the probability of the second star being blue .
18. R Bl B R R Bl B Bl R Bl B B 7 16 4 16 5 16 16 3 8 16 5 16 4 16 6 16 6 16 5 16 7 16 4 16 Note: look for the sequences where the second star is blue.
19. -Since we are looking for the probabilty of the second star being blue, we look at the tree diagram for the sequence where the second branch is blue. -There are 16 stars. The denominator of the probability fractions always stays 16 because the stars are being replaced. -Moving down the branch, you take the probability of each outcome in a sequence and multiply it by each other. -You the take that total and add it to the total of the other sequences that have the second star as blue.
20. Calculating the probability of the second star being blue... P ( R , B ) P (Bl, B ) P ( B , B ) = = = = = = 4 ( ) 16 5 16 ( ) 256 20 ( ) 7 16 ( ) 6 16 ( ) 42 256 ( ) 16 5 ( ) 4 16 ( ) 256 20 ( )
21. = 82 256 = 0.3203 = 32.0% There is a 32% probability that the second star will be blue. = 256 20 ( ) 42 256 ( ) 256 20 ( ) + + 41 128