Probability with one percentage
up vote
-2
down vote
favorite
If 40% of people have watched TV today, what is the probability that the three people you pick haven't watched TV today? The people are randomly selected and not replaced. I've just been stumped on this problem even though I know I've done something similar.
statistics
asked Nov 18 at 23:13
Chloe Hardin
62
add a comment |
up vote
-2
down vote
favorite
If 40% of people have watched TV today, what is the probability that the three people you pick haven't watched TV today? The people are randomly selected and not replaced. I've just been stumped on this problem even though I know I've done something similar.
statistics
asked Nov 18 at 23:13
Chloe Hardin
62
What are your thoughts on it?
– WaveX
Nov 18 at 23:17
How are you picking the people? Even assuming equal probability in picking, you could pick with replacement or without.
– Matt Samuel
Nov 18 at 23:19
We hope to see what you have tried and to know why you are not able to answer on your own. Please edit your Question to include your thoughts so far. // If you know about the binomial distribution, then let $X sim mathsf{Binom}(3, .6).$ You seek $P(X=3).$
– BruceET
Nov 18 at 23:20
The people are randomly selected and they are not replaced
– Chloe Hardin
Nov 18 at 23:20
If $40%$ of people watch TV, then what percentage of people didn't?
– WaveX
Nov 18 at 23:26
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
If 40% of people have watched TV today, what is the probability that the three people you pick haven't watched TV today? The people are randomly selected and not replaced. I've just been stumped on this problem even though I know I've done something similar.
statistics
asked Nov 18 at 23:13
Chloe Hardin
62
If 40% of people have watched TV today, what is the probability that the three people you pick haven't watched TV today? The people are randomly selected and not replaced. I've just been stumped on this problem even though I know I've done something similar.
statistics
statistics
asked Nov 18 at 23:13
Chloe Hardin
62
asked Nov 18 at 23:13
Chloe Hardin
62
edited Nov 18 at 23:23
asked Nov 18 at 23:13
Chloe Hardin
62
asked Nov 18 at 23:13
Chloe Hardin
62
asked Nov 18 at 23:13
Chloe Hardin
62
62
What are your thoughts on it?
– WaveX
Nov 18 at 23:17
How are you picking the people? Even assuming equal probability in picking, you could pick with replacement or without.
– Matt Samuel
Nov 18 at 23:19
We hope to see what you have tried and to know why you are not able to answer on your own. Please edit your Question to include your thoughts so far. // If you know about the binomial distribution, then let $X sim mathsf{Binom}(3, .6).$ You seek $P(X=3).$
– BruceET
Nov 18 at 23:20
The people are randomly selected and they are not replaced
– Chloe Hardin
Nov 18 at 23:20
If $40%$ of people watch TV, then what percentage of people didn't?
– WaveX
Nov 18 at 23:26
add a comment |
What are your thoughts on it?
– WaveX
Nov 18 at 23:17
How are you picking the people? Even assuming equal probability in picking, you could pick with replacement or without.
– Matt Samuel
Nov 18 at 23:19
We hope to see what you have tried and to know why you are not able to answer on your own. Please edit your Question to include your thoughts so far. // If you know about the binomial distribution, then let $X sim mathsf{Binom}(3, .6).$ You seek $P(X=3).$
– BruceET
Nov 18 at 23:20
The people are randomly selected and they are not replaced
– Chloe Hardin
Nov 18 at 23:20
If $40%$ of people watch TV, then what percentage of people didn't?
– WaveX
Nov 18 at 23:26
What are your thoughts on it?
– WaveX
Nov 18 at 23:17
What are your thoughts on it?
– WaveX
Nov 18 at 23:17
How are you picking the people? Even assuming equal probability in picking, you could pick with replacement or without.
– Matt Samuel
Nov 18 at 23:19
How are you picking the people? Even assuming equal probability in picking, you could pick with replacement or without.
– Matt Samuel
Nov 18 at 23:19
We hope to see what you have tried and to know why you are not able to answer on your own. Please edit your Question to include your thoughts so far. // If you know about the binomial distribution, then let $X sim mathsf{Binom}(3, .6).$ You seek $P(X=3).$
– BruceET
Nov 18 at 23:20
We hope to see what you have tried and to know why you are not able to answer on your own. Please edit your Question to include your thoughts so far. // If you know about the binomial distribution, then let $X sim mathsf{Binom}(3, .6).$ You seek $P(X=3).$
– BruceET
Nov 18 at 23:20
The people are randomly selected and they are not replaced
– Chloe Hardin
Nov 18 at 23:20
The people are randomly selected and they are not replaced
– Chloe Hardin
Nov 18 at 23:20
If $40%$ of people watch TV, then what percentage of people didn't?
– WaveX
Nov 18 at 23:26
If $40%$ of people watch TV, then what percentage of people didn't?
– WaveX
Nov 18 at 23:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
P(TV) = 0.4
P(Not TV) = 0.6
0.6 * 0.6 * 0.6 = 0.216 OR 21.6%
This is assuming that it is a complementary scenario.
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
add a comment |
up vote
0
down vote
Let A= watching tv. B= not watching tv.
$\P(B_{1} cap B_{2}cap B_{3})$ are all independent events. So you can use the multiplication rule.
$\P(B_{1} cap B_{2}cap B_{3}) = (P(B))^{3}=(0.6)^3$ It is important to know why you can do that.
answered Nov 19 at 0:20
pfmr1995
113
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
P(TV) = 0.4
P(Not TV) = 0.6
0.6 * 0.6 * 0.6 = 0.216 OR 21.6%
This is assuming that it is a complementary scenario.
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
add a comment |
up vote
0
down vote
accepted
P(TV) = 0.4
P(Not TV) = 0.6
0.6 * 0.6 * 0.6 = 0.216 OR 21.6%
This is assuming that it is a complementary scenario.
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
P(TV) = 0.4
P(Not TV) = 0.6
0.6 * 0.6 * 0.6 = 0.216 OR 21.6%
This is assuming that it is a complementary scenario.
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
P(TV) = 0.4
P(Not TV) = 0.6
0.6 * 0.6 * 0.6 = 0.216 OR 21.6%
This is assuming that it is a complementary scenario.
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
answered Nov 18 at 23:29
THIS_GAMES_DOODOO
212
212
add a comment |
add a comment |
up vote
0
down vote
Let A= watching tv. B= not watching tv.
$\P(B_{1} cap B_{2}cap B_{3})$ are all independent events. So you can use the multiplication rule.
$\P(B_{1} cap B_{2}cap B_{3}) = (P(B))^{3}=(0.6)^3$ It is important to know why you can do that.
answered Nov 19 at 0:20
pfmr1995
113
add a comment |
up vote
0
down vote
Let A= watching tv. B= not watching tv.
$\P(B_{1} cap B_{2}cap B_{3})$ are all independent events. So you can use the multiplication rule.
$\P(B_{1} cap B_{2}cap B_{3}) = (P(B))^{3}=(0.6)^3$ It is important to know why you can do that.
answered Nov 19 at 0:20
pfmr1995
113
add a comment |
up vote
0
down vote
up vote
0
down vote
Let A= watching tv. B= not watching tv.
$\P(B_{1} cap B_{2}cap B_{3})$ are all independent events. So you can use the multiplication rule.
$\P(B_{1} cap B_{2}cap B_{3}) = (P(B))^{3}=(0.6)^3$ It is important to know why you can do that.
answered Nov 19 at 0:20
pfmr1995
113
Let A= watching tv. B= not watching tv.
$\P(B_{1} cap B_{2}cap B_{3})$ are all independent events. So you can use the multiplication rule.
$\P(B_{1} cap B_{2}cap B_{3}) = (P(B))^{3}=(0.6)^3$ It is important to know why you can do that.
answered Nov 19 at 0:20
pfmr1995
113
answered Nov 19 at 0:20
pfmr1995
113
answered Nov 19 at 0:20
pfmr1995
113
answered Nov 19 at 0:20
pfmr1995
113
113
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3004265%2fprobability-with-one-percentage%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
What are your thoughts on it?
– WaveX
Nov 18 at 23:17
How are you picking the people? Even assuming equal probability in picking, you could pick with replacement or without.
– Matt Samuel
Nov 18 at 23:19
We hope to see what you have tried and to know why you are not able to answer on your own. Please edit your Question to include your thoughts so far. // If you know about the binomial distribution, then let $X sim mathsf{Binom}(3, .6).$ You seek $P(X=3).$
– BruceET
Nov 18 at 23:20
The people are randomly selected and they are not replaced
– Chloe Hardin
Nov 18 at 23:20
If $40%$ of people watch TV, then what percentage of people didn't?
– WaveX
Nov 18 at 23:26