Mixing
Probability of drawing all six numbers different in Powerball lottery
$begingroup$
I came across the following problem involving the Powerball lottery.
The Powerball lottery has the following rules. There are $69$ white balls (numbered from $1$ to $69$) and $26$ red balls (numbered from $1$ to $26$). At each drawing five white balls and a red ball are chosen uniformly at random. The result of the drawing is the five white numbers are listed in decreasing order and the red number.
Find the probability that the picked six numbers are all different.
So far I have that the number of ways to pick the first $5$ white balls is equal to $69cdot68cdot67cdot66cdot65.$ However I am having trouble understanding how to account for the red powerball, which has a different range of possible outcomes, being $1$ to $26.$ So that when if all five white balls are chosen over $27$ for instance, we still have $26$ choices for the red powerball, but if all five white ball are chosen below $26,$ now we have only $21$ choices for the powerball.
Any hints on how to proceed are greatly appreciated.
probability
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
$endgroup$
add a comment |
$begingroup$
I came across the following problem involving the Powerball lottery.
The Powerball lottery has the following rules. There are $69$ white balls (numbered from $1$ to $69$) and $26$ red balls (numbered from $1$ to $26$). At each drawing five white balls and a red ball are chosen uniformly at random. The result of the drawing is the five white numbers are listed in decreasing order and the red number.
Find the probability that the picked six numbers are all different.
So far I have that the number of ways to pick the first $5$ white balls is equal to $69cdot68cdot67cdot66cdot65.$ However I am having trouble understanding how to account for the red powerball, which has a different range of possible outcomes, being $1$ to $26.$ So that when if all five white balls are chosen over $27$ for instance, we still have $26$ choices for the red powerball, but if all five white ball are chosen below $26,$ now we have only $21$ choices for the powerball.
Any hints on how to proceed are greatly appreciated.
probability
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
$endgroup$
add a comment |
$begingroup$
I came across the following problem involving the Powerball lottery.
The Powerball lottery has the following rules. There are $69$ white balls (numbered from $1$ to $69$) and $26$ red balls (numbered from $1$ to $26$). At each drawing five white balls and a red ball are chosen uniformly at random. The result of the drawing is the five white numbers are listed in decreasing order and the red number.
Find the probability that the picked six numbers are all different.
So far I have that the number of ways to pick the first $5$ white balls is equal to $69cdot68cdot67cdot66cdot65.$ However I am having trouble understanding how to account for the red powerball, which has a different range of possible outcomes, being $1$ to $26.$ So that when if all five white balls are chosen over $27$ for instance, we still have $26$ choices for the red powerball, but if all five white ball are chosen below $26,$ now we have only $21$ choices for the powerball.
Any hints on how to proceed are greatly appreciated.
probability
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
$endgroup$
I came across the following problem involving the Powerball lottery.
The Powerball lottery has the following rules. There are $69$ white balls (numbered from $1$ to $69$) and $26$ red balls (numbered from $1$ to $26$). At each drawing five white balls and a red ball are chosen uniformly at random. The result of the drawing is the five white numbers are listed in decreasing order and the red number.
Find the probability that the picked six numbers are all different.
So far I have that the number of ways to pick the first $5$ white balls is equal to $69cdot68cdot67cdot66cdot65.$ However I am having trouble understanding how to account for the red powerball, which has a different range of possible outcomes, being $1$ to $26.$ So that when if all five white balls are chosen over $27$ for instance, we still have $26$ choices for the red powerball, but if all five white ball are chosen below $26,$ now we have only $21$ choices for the powerball.
Any hints on how to proceed are greatly appreciated.
probability
probability
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
edited Jan 31 at 11:33
Gaby Alfonso
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
asked Jan 31 at 11:05
Gaby AlfonsoGaby Alfonso
1,1931418
1,1931418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This question can be done by decomposing the event into 2 parts:
The thing about decreasing order is completely irrelevant to the actual solution.
Part 1: Choosing a red ball. There are 26 choices.
Part 2: Choosing the 5 white balls. This is $68choose5$. We set 68 because we want to avoid the ball we already picked( the red one)
Multiply them together.
Divide by the total amount of choices, which is 26 times $69choose5$. The result is your probability
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
$endgroup$
add a comment |
$begingroup$
Some hints
Consider that a coincidence can happen only if the red ball equals one of the white balls.
Compute the probability of coincidence given that (conditioned to) the red ball is $1$
Compute the probability of coincidence (total probability)
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
$endgroup$
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%2f3094767%2fprobability-of-drawing-all-six-numbers-different-in-powerball-lottery%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This question can be done by decomposing the event into 2 parts:
The thing about decreasing order is completely irrelevant to the actual solution.
Part 1: Choosing a red ball. There are 26 choices.
Part 2: Choosing the 5 white balls. This is $68choose5$. We set 68 because we want to avoid the ball we already picked( the red one)
Multiply them together.
Divide by the total amount of choices, which is 26 times $69choose5$. The result is your probability
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
$endgroup$
add a comment |
$begingroup$
This question can be done by decomposing the event into 2 parts:
The thing about decreasing order is completely irrelevant to the actual solution.
Part 1: Choosing a red ball. There are 26 choices.
Part 2: Choosing the 5 white balls. This is $68choose5$. We set 68 because we want to avoid the ball we already picked( the red one)
Multiply them together.
Divide by the total amount of choices, which is 26 times $69choose5$. The result is your probability
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
$endgroup$
add a comment |
$begingroup$
This question can be done by decomposing the event into 2 parts:
The thing about decreasing order is completely irrelevant to the actual solution.
Part 1: Choosing a red ball. There are 26 choices.
Part 2: Choosing the 5 white balls. This is $68choose5$. We set 68 because we want to avoid the ball we already picked( the red one)
Multiply them together.
Divide by the total amount of choices, which is 26 times $69choose5$. The result is your probability
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
$endgroup$
This question can be done by decomposing the event into 2 parts:
The thing about decreasing order is completely irrelevant to the actual solution.
Part 1: Choosing a red ball. There are 26 choices.
Part 2: Choosing the 5 white balls. This is $68choose5$. We set 68 because we want to avoid the ball we already picked( the red one)
Multiply them together.
Divide by the total amount of choices, which is 26 times $69choose5$. The result is your probability
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
answered Jan 31 at 11:17
Marat AlievMarat Aliev
1312
1312
add a comment |
add a comment |
$begingroup$
Some hints
Consider that a coincidence can happen only if the red ball equals one of the white balls.
Compute the probability of coincidence given that (conditioned to) the red ball is $1$
Compute the probability of coincidence (total probability)
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
$endgroup$
add a comment |
$begingroup$
Some hints
Consider that a coincidence can happen only if the red ball equals one of the white balls.
Compute the probability of coincidence given that (conditioned to) the red ball is $1$
Compute the probability of coincidence (total probability)
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
$endgroup$
add a comment |
$begingroup$
Some hints
Consider that a coincidence can happen only if the red ball equals one of the white balls.
Compute the probability of coincidence given that (conditioned to) the red ball is $1$
Compute the probability of coincidence (total probability)
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
$endgroup$
Some hints
Consider that a coincidence can happen only if the red ball equals one of the white balls.
Compute the probability of coincidence given that (conditioned to) the red ball is $1$
Compute the probability of coincidence (total probability)
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
answered Jan 31 at 11:12
leonbloyleonbloy
42.2k647108
42.2k647108
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%2f3094767%2fprobability-of-drawing-all-six-numbers-different-in-powerball-lottery%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
