Mixing
mean of buying different colored candy(tricky)
$begingroup$
i am having a problem with the following problem:
a guy loves his girlfriend, so he buys her each day a new bag of colored candies. there's no dependancy between bags of candy in different days.
a)what is the mean of number of candy bags he'll buy, if in each bag there are exactly two candies, and each candy can be in one of three colors: red, green and blue, with the probability of red:1/8, green:1/4, blue: 5/8, and the guy will keep buying candy bags until she gets at least 2 green candies?
b)what is the mean of the number of colors of candy that his girlfriend will have, if he buys exactly 10 bags, and in each bag there's only one candy, in one of 6 colors, in same probability?
here's what i tried, even though i don't know exactly how to continue:
a)let X be the number of candy bags he'll buy until there are at least 2 green candies. $X_1+X_2=X$, where $X_1$ will be the number of bags he bought and there's one green candy, and $X_2$ the amount of bags he'll have to buy until obtaining 2 green candies. so $E(X) = E(x_1) + E(x_2)$, and for $x_1$ he needs to keep buying bags until the first success, so it's a geometric distribution, so it's $x_1 ~ geo(1-0.25=0.75)$. then the mean is $E(x_1) = 1/p = frac{4}{3}$. next it's the second part of the experiment, where he continues buying until he gets the second one is the same, we need to continue until we get the second one so $x_4 ~ geo(1-1/4=3/4)$. from here i don't know how to continue or if what i did is correct.
b) for each of the 6 possible colors, $X_i$ is defined to be 1 if the colors i was obtained once, or 0 otherwise. thus, $X=sum_{i=1}^6X_i$ is the number of 6 colored candies in 10 bags. so i need to find $p(x_i=1)=1-p(x_i=0)=1-frac{5}{6}^{10}=0.8385$ so $E[x]=E[sum_{i=1}^6x_i]=sum_{i=1}^6E[x_i]=6*0.8385$. is this correct?
hoping that what i did so far is correct, would appreciate learning what i did wrong and how to improve. thank you very much for your help!
probability probability-distributions problem-solving
asked Jan 29 at 19:30
q123q123
75
$endgroup$
add a comment |
$begingroup$
i am having a problem with the following problem:
a guy loves his girlfriend, so he buys her each day a new bag of colored candies. there's no dependancy between bags of candy in different days.
a)what is the mean of number of candy bags he'll buy, if in each bag there are exactly two candies, and each candy can be in one of three colors: red, green and blue, with the probability of red:1/8, green:1/4, blue: 5/8, and the guy will keep buying candy bags until she gets at least 2 green candies?
b)what is the mean of the number of colors of candy that his girlfriend will have, if he buys exactly 10 bags, and in each bag there's only one candy, in one of 6 colors, in same probability?
here's what i tried, even though i don't know exactly how to continue:
a)let X be the number of candy bags he'll buy until there are at least 2 green candies. $X_1+X_2=X$, where $X_1$ will be the number of bags he bought and there's one green candy, and $X_2$ the amount of bags he'll have to buy until obtaining 2 green candies. so $E(X) = E(x_1) + E(x_2)$, and for $x_1$ he needs to keep buying bags until the first success, so it's a geometric distribution, so it's $x_1 ~ geo(1-0.25=0.75)$. then the mean is $E(x_1) = 1/p = frac{4}{3}$. next it's the second part of the experiment, where he continues buying until he gets the second one is the same, we need to continue until we get the second one so $x_4 ~ geo(1-1/4=3/4)$. from here i don't know how to continue or if what i did is correct.
b) for each of the 6 possible colors, $X_i$ is defined to be 1 if the colors i was obtained once, or 0 otherwise. thus, $X=sum_{i=1}^6X_i$ is the number of 6 colored candies in 10 bags. so i need to find $p(x_i=1)=1-p(x_i=0)=1-frac{5}{6}^{10}=0.8385$ so $E[x]=E[sum_{i=1}^6x_i]=sum_{i=1}^6E[x_i]=6*0.8385$. is this correct?
hoping that what i did so far is correct, would appreciate learning what i did wrong and how to improve. thank you very much for your help!
probability probability-distributions problem-solving
asked Jan 29 at 19:30
q123q123
75
$endgroup$
$begingroup$
You are correct on (b). For (a), you might take a look at the negative binomial distribution. en.wikipedia.org/wiki/Negative_binomial_distribution
$endgroup$
– awkward
Feb 7 at 14:37
add a comment |
$begingroup$
i am having a problem with the following problem:
a guy loves his girlfriend, so he buys her each day a new bag of colored candies. there's no dependancy between bags of candy in different days.
a)what is the mean of number of candy bags he'll buy, if in each bag there are exactly two candies, and each candy can be in one of three colors: red, green and blue, with the probability of red:1/8, green:1/4, blue: 5/8, and the guy will keep buying candy bags until she gets at least 2 green candies?
b)what is the mean of the number of colors of candy that his girlfriend will have, if he buys exactly 10 bags, and in each bag there's only one candy, in one of 6 colors, in same probability?
here's what i tried, even though i don't know exactly how to continue:
a)let X be the number of candy bags he'll buy until there are at least 2 green candies. $X_1+X_2=X$, where $X_1$ will be the number of bags he bought and there's one green candy, and $X_2$ the amount of bags he'll have to buy until obtaining 2 green candies. so $E(X) = E(x_1) + E(x_2)$, and for $x_1$ he needs to keep buying bags until the first success, so it's a geometric distribution, so it's $x_1 ~ geo(1-0.25=0.75)$. then the mean is $E(x_1) = 1/p = frac{4}{3}$. next it's the second part of the experiment, where he continues buying until he gets the second one is the same, we need to continue until we get the second one so $x_4 ~ geo(1-1/4=3/4)$. from here i don't know how to continue or if what i did is correct.
b) for each of the 6 possible colors, $X_i$ is defined to be 1 if the colors i was obtained once, or 0 otherwise. thus, $X=sum_{i=1}^6X_i$ is the number of 6 colored candies in 10 bags. so i need to find $p(x_i=1)=1-p(x_i=0)=1-frac{5}{6}^{10}=0.8385$ so $E[x]=E[sum_{i=1}^6x_i]=sum_{i=1}^6E[x_i]=6*0.8385$. is this correct?
hoping that what i did so far is correct, would appreciate learning what i did wrong and how to improve. thank you very much for your help!
probability probability-distributions problem-solving
asked Jan 29 at 19:30
q123q123
75
$endgroup$
i am having a problem with the following problem:
a guy loves his girlfriend, so he buys her each day a new bag of colored candies. there's no dependancy between bags of candy in different days.
a)what is the mean of number of candy bags he'll buy, if in each bag there are exactly two candies, and each candy can be in one of three colors: red, green and blue, with the probability of red:1/8, green:1/4, blue: 5/8, and the guy will keep buying candy bags until she gets at least 2 green candies?
b)what is the mean of the number of colors of candy that his girlfriend will have, if he buys exactly 10 bags, and in each bag there's only one candy, in one of 6 colors, in same probability?
here's what i tried, even though i don't know exactly how to continue:
a)let X be the number of candy bags he'll buy until there are at least 2 green candies. $X_1+X_2=X$, where $X_1$ will be the number of bags he bought and there's one green candy, and $X_2$ the amount of bags he'll have to buy until obtaining 2 green candies. so $E(X) = E(x_1) + E(x_2)$, and for $x_1$ he needs to keep buying bags until the first success, so it's a geometric distribution, so it's $x_1 ~ geo(1-0.25=0.75)$. then the mean is $E(x_1) = 1/p = frac{4}{3}$. next it's the second part of the experiment, where he continues buying until he gets the second one is the same, we need to continue until we get the second one so $x_4 ~ geo(1-1/4=3/4)$. from here i don't know how to continue or if what i did is correct.
b) for each of the 6 possible colors, $X_i$ is defined to be 1 if the colors i was obtained once, or 0 otherwise. thus, $X=sum_{i=1}^6X_i$ is the number of 6 colored candies in 10 bags. so i need to find $p(x_i=1)=1-p(x_i=0)=1-frac{5}{6}^{10}=0.8385$ so $E[x]=E[sum_{i=1}^6x_i]=sum_{i=1}^6E[x_i]=6*0.8385$. is this correct?
hoping that what i did so far is correct, would appreciate learning what i did wrong and how to improve. thank you very much for your help!
probability probability-distributions problem-solving
probability probability-distributions problem-solving
asked Jan 29 at 19:30
q123q123
75
asked Jan 29 at 19:30
q123q123
75
asked Jan 29 at 19:30
q123q123
75
asked Jan 29 at 19:30
q123q123
75
asked Jan 29 at 19:30
q123q123
75
75
$begingroup$
You are correct on (b). For (a), you might take a look at the negative binomial distribution. en.wikipedia.org/wiki/Negative_binomial_distribution
$endgroup$
– awkward
Feb 7 at 14:37
add a comment |
$begingroup$
You are correct on (b). For (a), you might take a look at the negative binomial distribution. en.wikipedia.org/wiki/Negative_binomial_distribution
$endgroup$
– awkward
Feb 7 at 14:37
$begingroup$
You are correct on (b). For (a), you might take a look at the negative binomial distribution. en.wikipedia.org/wiki/Negative_binomial_distribution
$endgroup$
– awkward
Feb 7 at 14:37
$begingroup$
You are correct on (b). For (a), you might take a look at the negative binomial distribution. en.wikipedia.org/wiki/Negative_binomial_distribution
$endgroup$
– awkward
Feb 7 at 14:37
add a comment |
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$begingroup$
You are correct on (b). For (a), you might take a look at the negative binomial distribution. en.wikipedia.org/wiki/Negative_binomial_distribution
$endgroup$
– awkward
Feb 7 at 14:37