If two different integers between 1 and 100 inclusive are chosen at random, what is the probability that the difference of the two numbers is 15? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Probability depending on x being the difference between two numbersProbability between two numbers chosen from 1 to 30.Probability that all colors are chosentwo integers are chosen at random between $0$ and $10$ what is the probability that they differ by no more than $5$?What is the probability that $7$ cards are chosen and no suit is missing?Probability of difference 5 in pairs of random chosen numbersWhat is the probability that Jan and Jon are chosen?What is the probability that a randomly chosen integer between 0000 and 9999 is divisible by 5?Probability that people are chosen in different groups?Two numbers are chosen independently and at random.
Random body shuffle every night—can we still function?
How do you cope with tons of web fonts when copying and pasting from web pages?
Short story about astronauts fertilizing soil with their own bodies
Can anyone explain what's the meaning of this in the new Game of Thrones opening animations?
Why do C and C++ allow the expression (int) + 4*5?
Cost function for LTI system identification
What should one know about term logic before studying propositional and predicate logic?
Improvising over quartal voicings
What helicopter has the most rotor blades?
My mentor says to set image to Fine instead of RAW — how is this different from JPG?
How did 'ликвиди́ровать' semantically shift to mean 'abolish' and 'destroy, kill'?
Why not use the yoke to control yaw, as well as pitch and roll?
calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle
Why are current probes so expensive?
Are there any irrational/transcendental numbers for which the distribution of decimal digits is not uniform?
3D Masyu - A Die
RM anova or Factorial Anova?
How to get a flat-head nail out of a piece of wood?
Why complex landing gears are used instead of simple, reliable and light weight muscle wire or shape memory alloys?
Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?
Who's this lady in the war room?
How to resize main filesystem
Can I take recommendation from someone I met at a conference?
Is there a canonical “inverse” of Abelianization?
If two different integers between 1 and 100 inclusive are chosen at random, what is the probability that the difference of the two numbers is 15?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Probability depending on x being the difference between two numbersProbability between two numbers chosen from 1 to 30.Probability that all colors are chosentwo integers are chosen at random between $0$ and $10$ what is the probability that they differ by no more than $5$?What is the probability that $7$ cards are chosen and no suit is missing?Probability of difference 5 in pairs of random chosen numbersWhat is the probability that Jan and Jon are chosen?What is the probability that a randomly chosen integer between 0000 and 9999 is divisible by 5?Probability that people are chosen in different groups?Two numbers are chosen independently and at random.
$begingroup$
My work:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (1-20)
20-5
19-4
18-3
17-2
16-1 so 5
5^5=3125 Am I missing a step? Or am I correct?
calculus probability combinations
asked Apr 2 at 17:17
MaryMary
316
$endgroup$
add a comment |
$begingroup$
My work:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (1-20)
20-5
19-4
18-3
17-2
16-1 so 5
5^5=3125 Am I missing a step? Or am I correct?
calculus probability combinations
asked Apr 2 at 17:17
MaryMary
316
$endgroup$
1
$begingroup$
They ask for a probability, so you need a number between $0$ and $1$. Also, no idea why $5^5$ is relevant to your problem. And why did you try to solve it only for the 1-20 range?
$endgroup$
– Andrei
Apr 2 at 17:23
$begingroup$
There are some characters in your question that do not render properly. You need to add some words to explain what you are doing. Your question seems to work only up to $20$ while the title calls for $100$.
$endgroup$
– Ross Millikan
Apr 2 at 17:23
$begingroup$
You are missing the probability? What does $5^5$ represent?
$endgroup$
– Thomas Andrews
Apr 2 at 17:30
add a comment |
$begingroup$
My work:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (1-20)
20-5
19-4
18-3
17-2
16-1 so 5
5^5=3125 Am I missing a step? Or am I correct?
calculus probability combinations
asked Apr 2 at 17:17
MaryMary
316
$endgroup$
My work:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 (1-20)
20-5
19-4
18-3
17-2
16-1 so 5
5^5=3125 Am I missing a step? Or am I correct?
calculus probability combinations
calculus probability combinations
asked Apr 2 at 17:17
MaryMary
316
asked Apr 2 at 17:17
MaryMary
316
asked Apr 2 at 17:17
MaryMary
316
asked Apr 2 at 17:17
MaryMary
316
asked Apr 2 at 17:17
MaryMary
316
316
1
$begingroup$
They ask for a probability, so you need a number between $0$ and $1$. Also, no idea why $5^5$ is relevant to your problem. And why did you try to solve it only for the 1-20 range?
$endgroup$
– Andrei
Apr 2 at 17:23
$begingroup$
There are some characters in your question that do not render properly. You need to add some words to explain what you are doing. Your question seems to work only up to $20$ while the title calls for $100$.
$endgroup$
– Ross Millikan
Apr 2 at 17:23
$begingroup$
You are missing the probability? What does $5^5$ represent?
$endgroup$
– Thomas Andrews
Apr 2 at 17:30
add a comment |
1
$begingroup$
They ask for a probability, so you need a number between $0$ and $1$. Also, no idea why $5^5$ is relevant to your problem. And why did you try to solve it only for the 1-20 range?
$endgroup$
– Andrei
Apr 2 at 17:23
$begingroup$
There are some characters in your question that do not render properly. You need to add some words to explain what you are doing. Your question seems to work only up to $20$ while the title calls for $100$.
$endgroup$
– Ross Millikan
Apr 2 at 17:23
$begingroup$
You are missing the probability? What does $5^5$ represent?
$endgroup$
– Thomas Andrews
Apr 2 at 17:30
1
1
$begingroup$
They ask for a probability, so you need a number between $0$ and $1$. Also, no idea why $5^5$ is relevant to your problem. And why did you try to solve it only for the 1-20 range?
$endgroup$
– Andrei
Apr 2 at 17:23
$begingroup$
They ask for a probability, so you need a number between $0$ and $1$. Also, no idea why $5^5$ is relevant to your problem. And why did you try to solve it only for the 1-20 range?
$endgroup$
– Andrei
Apr 2 at 17:23
$begingroup$
There are some characters in your question that do not render properly. You need to add some words to explain what you are doing. Your question seems to work only up to $20$ while the title calls for $100$.
$endgroup$
– Ross Millikan
Apr 2 at 17:23
$begingroup$
There are some characters in your question that do not render properly. You need to add some words to explain what you are doing. Your question seems to work only up to $20$ while the title calls for $100$.
$endgroup$
– Ross Millikan
Apr 2 at 17:23
$begingroup$
You are missing the probability? What does $5^5$ represent?
$endgroup$
– Thomas Andrews
Apr 2 at 17:30
$begingroup$
You are missing the probability? What does $5^5$ represent?
$endgroup$
– Thomas Andrews
Apr 2 at 17:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: How many pairs have a difference of $15?$ How many pairs are there? Divide to get the probability.
Added: if the first number is $1-15$ or $86-100$, a $0.3$ probability, you have $frac 199$, while if the first number is $16-85$ you have $frac 299$, so the overall chance is $$0.3 cdot frac 199+0.7 cdot frac 299=frac 17990$$
Another way to see it is that there are $85$ pairs $15$ apart out of $100 choose 2=4950$ total pairs, so the chance is $$frac 854950=frac 17990$$
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
$endgroup$
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3172135%2fif-two-different-integers-between-1-and-100-inclusive-are-chosen-at-random-what%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: How many pairs have a difference of $15?$ How many pairs are there? Divide to get the probability.
Added: if the first number is $1-15$ or $86-100$, a $0.3$ probability, you have $frac 199$, while if the first number is $16-85$ you have $frac 299$, so the overall chance is $$0.3 cdot frac 199+0.7 cdot frac 299=frac 17990$$
Another way to see it is that there are $85$ pairs $15$ apart out of $100 choose 2=4950$ total pairs, so the chance is $$frac 854950=frac 17990$$
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
$endgroup$
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
add a comment |
$begingroup$
Hint: How many pairs have a difference of $15?$ How many pairs are there? Divide to get the probability.
Added: if the first number is $1-15$ or $86-100$, a $0.3$ probability, you have $frac 199$, while if the first number is $16-85$ you have $frac 299$, so the overall chance is $$0.3 cdot frac 199+0.7 cdot frac 299=frac 17990$$
Another way to see it is that there are $85$ pairs $15$ apart out of $100 choose 2=4950$ total pairs, so the chance is $$frac 854950=frac 17990$$
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
$endgroup$
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
add a comment |
$begingroup$
Hint: How many pairs have a difference of $15?$ How many pairs are there? Divide to get the probability.
Added: if the first number is $1-15$ or $86-100$, a $0.3$ probability, you have $frac 199$, while if the first number is $16-85$ you have $frac 299$, so the overall chance is $$0.3 cdot frac 199+0.7 cdot frac 299=frac 17990$$
Another way to see it is that there are $85$ pairs $15$ apart out of $100 choose 2=4950$ total pairs, so the chance is $$frac 854950=frac 17990$$
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
$endgroup$
Hint: How many pairs have a difference of $15?$ How many pairs are there? Divide to get the probability.
Added: if the first number is $1-15$ or $86-100$, a $0.3$ probability, you have $frac 199$, while if the first number is $16-85$ you have $frac 299$, so the overall chance is $$0.3 cdot frac 199+0.7 cdot frac 299=frac 17990$$
Another way to see it is that there are $85$ pairs $15$ apart out of $100 choose 2=4950$ total pairs, so the chance is $$frac 854950=frac 17990$$
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
edited Apr 2 at 22:57
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
answered Apr 2 at 17:25
Ross MillikanRoss Millikan
302k24201375
302k24201375
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
add a comment |
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
So it should be 85 as my answer?
$endgroup$
– Mary
Apr 2 at 17:52
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
That is the answer to the first question. You are asked for a probability, which must be between $0$ and $1$.
$endgroup$
– Ross Millikan
Apr 2 at 18:02
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
85/50 but it is not between 0 and 1 though. That means I did something wrong.
$endgroup$
– Mary
Apr 2 at 18:27
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
$begingroup$
There are many more pairs than $50$. There are $100 choose 2$. Intuitively, the answer has to be between $1/100$ and $2/100$ because some first cards you could draw only allow $1$ choice for the second card and some allow $2$.
$endgroup$
– Ross Millikan
Apr 2 at 18:39
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3172135%2fif-two-different-integers-between-1-and-100-inclusive-are-chosen-at-random-what%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
They ask for a probability, so you need a number between $0$ and $1$. Also, no idea why $5^5$ is relevant to your problem. And why did you try to solve it only for the 1-20 range?
$endgroup$
– Andrei
Apr 2 at 17:23
$begingroup$
There are some characters in your question that do not render properly. You need to add some words to explain what you are doing. Your question seems to work only up to $20$ while the title calls for $100$.
$endgroup$
– Ross Millikan
Apr 2 at 17:23
$begingroup$
You are missing the probability? What does $5^5$ represent?
$endgroup$
– Thomas Andrews
Apr 2 at 17:30