An integer is chosen at random from numbers 1 to 50, what is the probability that the integer chosen is a...
$begingroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
asked Jan 10 at 4:17
GarimaGarima
11
$endgroup$
add a comment |
$begingroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
asked Jan 10 at 4:17
GarimaGarima
11
$endgroup$
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
add a comment |
$begingroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
asked Jan 10 at 4:17
GarimaGarima
11
$endgroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
probability
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
asked Jan 10 at 4:17
GarimaGarima
11
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
asked Jan 10 at 4:17
GarimaGarima
11
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
edited Jan 10 at 4:24
Theo Bendit
20.3k12353
20.3k12353
asked Jan 10 at 4:17
GarimaGarima
11
asked Jan 10 at 4:17
GarimaGarima
11
asked Jan 10 at 4:17
GarimaGarima
11
11
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
add a comment |
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
4,611512
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3068233%2fan-integer-is-chosen-at-random-from-numbers-1-to-50-what-is-the-probability-tha%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
$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
4,611512
$endgroup$
add a comment |
$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
4,611512
$endgroup$
add a comment |
$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
4,611512
$endgroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
4,611512
answered Jan 10 at 4:32
MikeMike
4,611512
answered Jan 10 at 4:32
MikeMike
4,611512
answered Jan 10 at 4:32
MikeMike
4,611512
4,611512
add a comment |
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%2f3068233%2fan-integer-is-chosen-at-random-from-numbers-1-to-50-what-is-the-probability-tha%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
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29