Basic probability homework question help (combinatorical counting problem) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the problem in this solution to the Two Child Problem?Multinomial Theorem, distinct objects to identical boxes?Probability of having an empty locationConditional Probability QuestionHipergeometric. Kids and candy.A problem with too many counting requiredExtended Hypergeometric candy problemBasic probability question - CountingFind that probability that exactly $ 4 $ students get candyDistributing $15$ candies to $5$ children with restriction
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Basic probability homework question help (combinatorical counting problem)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the problem in this solution to the Two Child Problem?Multinomial Theorem, distinct objects to identical boxes?Probability of having an empty locationConditional Probability QuestionHipergeometric. Kids and candy.A problem with too many counting requiredExtended Hypergeometric candy problemBasic probability question - CountingFind that probability that exactly $ 4 $ students get candyDistributing $15$ candies to $5$ children with restriction
$begingroup$
I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes.
There are $6$ children in a family and many candies in $4$ colors - blue, red, green, pink. The $2$ eldest children get to choose $2$ candies each. The $4$ other children get to choose $1$ candy apiece. (The same candy type can be chosen by a few children).
a. Count the number of choices.
b. What is the probability that exactly $2$ children choose blue candies?
My attempt at solution:
a. $$binom42^2 cdot 4^4 = 6^2 cdot 4^4$$
This solution is correct according to the solutions given to me.
b.
I tried to solve it in the following way:
Dividing the question into 3 cases:
$(1)$ The number of cases in which the two eldest children choose blue candies:
$$3^4$$
$(2)$ The number of cases in which two younger children choose blue candies:
$$3^2 cdot 3^2 = 3^4$$
$(3)$ The number of cases in which one blue candy is chosen by an older child and one by a younger:
$$2 cdot 4 cdot 3 cdot 3^3 = 2 cdot 4 cdot 3^4$$
I sum all the case to get $$10 cdot 3^4$$
The final solution I get is: $$frac10 cdot 3^46^2 cdot 4^4 = 0.088$$
My issues with this question is that in the solutions given to me, the number of options in case $(1)$ is calculated as: $$3^2 cdot 3^4$$
and in case $(2)$: $$3^2 cdot binom42 cdot 3^2$$
My question is - why is this solution correct?
Any help would be greatly appreciated.
Thank you in advance!!
probability combinatorics
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
$endgroup$
add a comment |
$begingroup$
I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes.
There are $6$ children in a family and many candies in $4$ colors - blue, red, green, pink. The $2$ eldest children get to choose $2$ candies each. The $4$ other children get to choose $1$ candy apiece. (The same candy type can be chosen by a few children).
a. Count the number of choices.
b. What is the probability that exactly $2$ children choose blue candies?
My attempt at solution:
a. $$binom42^2 cdot 4^4 = 6^2 cdot 4^4$$
This solution is correct according to the solutions given to me.
b.
I tried to solve it in the following way:
Dividing the question into 3 cases:
$(1)$ The number of cases in which the two eldest children choose blue candies:
$$3^4$$
$(2)$ The number of cases in which two younger children choose blue candies:
$$3^2 cdot 3^2 = 3^4$$
$(3)$ The number of cases in which one blue candy is chosen by an older child and one by a younger:
$$2 cdot 4 cdot 3 cdot 3^3 = 2 cdot 4 cdot 3^4$$
I sum all the case to get $$10 cdot 3^4$$
The final solution I get is: $$frac10 cdot 3^46^2 cdot 4^4 = 0.088$$
My issues with this question is that in the solutions given to me, the number of options in case $(1)$ is calculated as: $$3^2 cdot 3^4$$
and in case $(2)$: $$3^2 cdot binom42 cdot 3^2$$
My question is - why is this solution correct?
Any help would be greatly appreciated.
Thank you in advance!!
probability combinatorics
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
$endgroup$
add a comment |
$begingroup$
I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes.
There are $6$ children in a family and many candies in $4$ colors - blue, red, green, pink. The $2$ eldest children get to choose $2$ candies each. The $4$ other children get to choose $1$ candy apiece. (The same candy type can be chosen by a few children).
a. Count the number of choices.
b. What is the probability that exactly $2$ children choose blue candies?
My attempt at solution:
a. $$binom42^2 cdot 4^4 = 6^2 cdot 4^4$$
This solution is correct according to the solutions given to me.
b.
I tried to solve it in the following way:
Dividing the question into 3 cases:
$(1)$ The number of cases in which the two eldest children choose blue candies:
$$3^4$$
$(2)$ The number of cases in which two younger children choose blue candies:
$$3^2 cdot 3^2 = 3^4$$
$(3)$ The number of cases in which one blue candy is chosen by an older child and one by a younger:
$$2 cdot 4 cdot 3 cdot 3^3 = 2 cdot 4 cdot 3^4$$
I sum all the case to get $$10 cdot 3^4$$
The final solution I get is: $$frac10 cdot 3^46^2 cdot 4^4 = 0.088$$
My issues with this question is that in the solutions given to me, the number of options in case $(1)$ is calculated as: $$3^2 cdot 3^4$$
and in case $(2)$: $$3^2 cdot binom42 cdot 3^2$$
My question is - why is this solution correct?
Any help would be greatly appreciated.
Thank you in advance!!
probability combinatorics
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
$endgroup$
I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes.
There are $6$ children in a family and many candies in $4$ colors - blue, red, green, pink. The $2$ eldest children get to choose $2$ candies each. The $4$ other children get to choose $1$ candy apiece. (The same candy type can be chosen by a few children).
a. Count the number of choices.
b. What is the probability that exactly $2$ children choose blue candies?
My attempt at solution:
a. $$binom42^2 cdot 4^4 = 6^2 cdot 4^4$$
This solution is correct according to the solutions given to me.
b.
I tried to solve it in the following way:
Dividing the question into 3 cases:
$(1)$ The number of cases in which the two eldest children choose blue candies:
$$3^4$$
$(2)$ The number of cases in which two younger children choose blue candies:
$$3^2 cdot 3^2 = 3^4$$
$(3)$ The number of cases in which one blue candy is chosen by an older child and one by a younger:
$$2 cdot 4 cdot 3 cdot 3^3 = 2 cdot 4 cdot 3^4$$
I sum all the case to get $$10 cdot 3^4$$
The final solution I get is: $$frac10 cdot 3^46^2 cdot 4^4 = 0.088$$
My issues with this question is that in the solutions given to me, the number of options in case $(1)$ is calculated as: $$3^2 cdot 3^4$$
and in case $(2)$: $$3^2 cdot binom42 cdot 3^2$$
My question is - why is this solution correct?
Any help would be greatly appreciated.
Thank you in advance!!
probability combinatorics
probability combinatorics
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
edited Mar 31 at 18:32
N. F. Taussig
45.3k103358
45.3k103358
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
asked Mar 31 at 17:56
PhysicsPrincessPhysicsPrincess
506
506
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the case 2 you must also consider the number of ways in which we can select 2 kids from the younger ones that is:
$$4choose 2$$
:)
answered Mar 31 at 18:22
EurekaEureka
876114
$endgroup$
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
add a comment |
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1 Answer
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$begingroup$
In the case 2 you must also consider the number of ways in which we can select 2 kids from the younger ones that is:
$$4choose 2$$
:)
answered Mar 31 at 18:22
EurekaEureka
876114
$endgroup$
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
add a comment |
$begingroup$
In the case 2 you must also consider the number of ways in which we can select 2 kids from the younger ones that is:
$$4choose 2$$
:)
answered Mar 31 at 18:22
EurekaEureka
876114
$endgroup$
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
add a comment |
$begingroup$
In the case 2 you must also consider the number of ways in which we can select 2 kids from the younger ones that is:
$$4choose 2$$
:)
answered Mar 31 at 18:22
EurekaEureka
876114
$endgroup$
In the case 2 you must also consider the number of ways in which we can select 2 kids from the younger ones that is:
$$4choose 2$$
:)
answered Mar 31 at 18:22
EurekaEureka
876114
answered Mar 31 at 18:22
EurekaEureka
876114
answered Mar 31 at 18:22
EurekaEureka
876114
answered Mar 31 at 18:22
EurekaEureka
876114
876114
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
add a comment |
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
$begingroup$
Oh sorry I missed that, thank you!
$endgroup$
– PhysicsPrincess
Mar 31 at 18:27
add a comment |
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