What's the probability of selecting a yellow skittle on second draw given that first skittle was green
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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 Nov 19 at 19:47
Drew Ngo
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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 Nov 19 at 19:47
Drew Ngo
1
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
Nov 19 at 19:55
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
Nov 19 at 20:04
@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
Nov 19 at 20:19
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ 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
Nov 19 at 20:37
@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
Nov 19 at 21:11
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up vote
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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 Nov 19 at 19:47
Drew Ngo
1
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 Nov 19 at 19:47
Drew Ngo
1
asked Nov 19 at 19:47
Drew Ngo
1
asked Nov 19 at 19:47
Drew Ngo
1
asked Nov 19 at 19:47
Drew Ngo
1
asked Nov 19 at 19:47
Drew Ngo
1
1
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
Nov 19 at 19:55
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
Nov 19 at 20:04
@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
Nov 19 at 20:19
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ 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
Nov 19 at 20:37
@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
Nov 19 at 21:11
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
Nov 19 at 19:55
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
Nov 19 at 20:04
@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
Nov 19 at 20:19
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ 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
Nov 19 at 20:37
@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
Nov 19 at 21:11
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
Nov 19 at 19:55
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
Nov 19 at 19:55
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
Nov 19 at 20:04
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
Nov 19 at 20:04
@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
Nov 19 at 20:19
@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
Nov 19 at 20:19
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ 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
Nov 19 at 20:37
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ 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
Nov 19 at 20:37
@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
Nov 19 at 21:11
@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
Nov 19 at 21:11
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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
Nov 19 at 19:55
However I interpret the exercise I never get a valid solution. Pls check the exercise
– callculus
Nov 19 at 20:04
@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
Nov 19 at 20:19
$P(G_1 cap Y_2) = P(G_1)P(Y_2|G_1).$ 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
Nov 19 at 20:37
@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
Nov 19 at 21:11