Mixing
What's the probability of selecting a yellow skittle on second draw given that first skittle was green
up vote
-1
down vote
favorite
A bag of skittles contains only 2 colors: green and yellow. Two skittles are randomly chosen without replacement. The probability of selecting a green and yellow skittle is 19/72 and the probability of selecting a green skittle on the first draw is 4/9. What's the probability of selecting a yellow skittle on the second draw given that the first skittle selected was green?
statistics
asked 2 days ago
Drew Ngo
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
-1
down vote
favorite
A bag of skittles contains only 2 colors: green and yellow. Two skittles are randomly chosen without replacement. The probability of selecting a green and yellow skittle is 19/72 and the probability of selecting a green skittle on the first draw is 4/9. What's the probability of selecting a yellow skittle on the second draw given that the first skittle selected was green?
statistics
asked 2 days ago
Drew Ngo
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
2 days ago
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
2 days ago
@callculus I seem to be getting $-frac{76}{149}$ green skittles in the bag and $-frac{95}{149}$ yellow skittles, giving $-frac{171}{149}$ skittles in total
– Henry
2 days ago
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ We are given $P(G_1) = 4/9.$ From the problem, it is unclear whether 'probability of green and yellow is 19/72' refers to $P(G_1 cap Y_2)$ or to $P(G_1 cap Y_2)+P(G_2 cap Y_1).$
– BruceET
2 days ago
@Henry It is not worth to think about it since the OP is not really interested in working on the problem. But I agree to your doubts.
– callculus
2 days ago
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
A bag of skittles contains only 2 colors: green and yellow. Two skittles are randomly chosen without replacement. The probability of selecting a green and yellow skittle is 19/72 and the probability of selecting a green skittle on the first draw is 4/9. What's the probability of selecting a yellow skittle on the second draw given that the first skittle selected was green?
statistics
asked 2 days ago
Drew Ngo
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A bag of skittles contains only 2 colors: green and yellow. Two skittles are randomly chosen without replacement. The probability of selecting a green and yellow skittle is 19/72 and the probability of selecting a green skittle on the first draw is 4/9. What's the probability of selecting a yellow skittle on the second draw given that the first skittle selected was green?
statistics
statistics
asked 2 days ago
Drew Ngo
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Drew Ngo
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Drew Ngo
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Drew Ngo
1
asked 2 days ago
Drew Ngo
1
1
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Drew Ngo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
2 days ago
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
2 days ago
@callculus I seem to be getting $-frac{76}{149}$ green skittles in the bag and $-frac{95}{149}$ yellow skittles, giving $-frac{171}{149}$ skittles in total
– Henry
2 days ago
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ We are given $P(G_1) = 4/9.$ From the problem, it is unclear whether 'probability of green and yellow is 19/72' refers to $P(G_1 cap Y_2)$ or to $P(G_1 cap Y_2)+P(G_2 cap Y_1).$
– BruceET
2 days ago
@Henry It is not worth to think about it since the OP is not really interested in working on the problem. But I agree to your doubts.
– callculus
2 days ago
add a comment |
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
2 days ago
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
2 days ago
@callculus I seem to be getting $-frac{76}{149}$ green skittles in the bag and $-frac{95}{149}$ yellow skittles, giving $-frac{171}{149}$ skittles in total
– Henry
2 days ago
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ We are given $P(G_1) = 4/9.$ From the problem, it is unclear whether 'probability of green and yellow is 19/72' refers to $P(G_1 cap Y_2)$ or to $P(G_1 cap Y_2)+P(G_2 cap Y_1).$
– BruceET
2 days ago
@Henry It is not worth to think about it since the OP is not really interested in working on the problem. But I agree to your doubts.
– callculus
2 days ago
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
2 days ago
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
2 days ago
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
2 days ago
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
2 days ago
@callculus I seem to be getting $-frac{76}{149}$ green skittles in the bag and $-frac{95}{149}$ yellow skittles, giving $-frac{171}{149}$ skittles in total
– Henry
2 days ago
@callculus I seem to be getting $-frac{76}{149}$ green skittles in the bag and $-frac{95}{149}$ yellow skittles, giving $-frac{171}{149}$ skittles in total
– Henry
2 days ago
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ We are given $P(G_1) = 4/9.$ From the problem, it is unclear whether 'probability of green and yellow is 19/72' refers to $P(G_1 cap Y_2)$ or to $P(G_1 cap Y_2)+P(G_2 cap Y_1).$
– BruceET
2 days ago
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ We are given $P(G_1) = 4/9.$ From the problem, it is unclear whether 'probability of green and yellow is 19/72' refers to $P(G_1 cap Y_2)$ or to $P(G_1 cap Y_2)+P(G_2 cap Y_1).$
– BruceET
2 days ago
@Henry It is not worth to think about it since the OP is not really interested in working on the problem. But I agree to your doubts.
– callculus
2 days ago
@Henry It is not worth to think about it since the OP is not really interested in working on the problem. But I agree to your doubts.
– callculus
2 days ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Drew Ngo is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Drew Ngo is a new contributor. Be nice, and check out our Code of Conduct.
Drew Ngo is a new contributor. Be nice, and check out our Code of Conduct.
Drew Ngo is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
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%2f3005424%2fwhats-the-probability-of-selecting-a-yellow-skittle-on-second-draw-given-that-f%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
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
2 days ago
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
2 days ago
@callculus I seem to be getting $-frac{76}{149}$ green skittles in the bag and $-frac{95}{149}$ yellow skittles, giving $-frac{171}{149}$ skittles in total
– Henry
2 days ago
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ We are given $P(G_1) = 4/9.$ From the problem, it is unclear whether 'probability of green and yellow is 19/72' refers to $P(G_1 cap Y_2)$ or to $P(G_1 cap Y_2)+P(G_2 cap Y_1).$
– BruceET
2 days ago
@Henry It is not worth to think about it since the OP is not really interested in working on the problem. But I agree to your doubts.
– callculus
2 days ago