Mixing
Probability that exactly 2 balls are white
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = frac36 = frac12$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,
$$ P(X=2) = 4 - 1 choose 2 - 1 left( frac12 right)^2 left( frac12 right)^2 $$
Which gives
$$ P(X=2) = boxeddfrac316 $$
Am I interpreting the problem correctly?
probability
edited 30 mins ago
Key Flex
5,450828
asked 1 hour ago
Neymar
30312
add a comment |Â
up vote
2
down vote
favorite
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = frac36 = frac12$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,
$$ P(X=2) = 4 - 1 choose 2 - 1 left( frac12 right)^2 left( frac12 right)^2 $$
Which gives
$$ P(X=2) = boxeddfrac316 $$
Am I interpreting the problem correctly?
probability
edited 30 mins ago
Key Flex
5,450828
asked 1 hour ago
Neymar
30312
$LaTeX$ Tip: use Bigr( frac 12 Bigr) to get $Bigr( frac 12 Bigr)$
– Mohammad Zuhair Khan
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = frac36 = frac12$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,
$$ P(X=2) = 4 - 1 choose 2 - 1 left( frac12 right)^2 left( frac12 right)^2 $$
Which gives
$$ P(X=2) = boxeddfrac316 $$
Am I interpreting the problem correctly?
probability
edited 30 mins ago
Key Flex
5,450828
asked 1 hour ago
Neymar
30312
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = frac36 = frac12$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,
$$ P(X=2) = 4 - 1 choose 2 - 1 left( frac12 right)^2 left( frac12 right)^2 $$
Which gives
$$ P(X=2) = boxeddfrac316 $$
Am I interpreting the problem correctly?
probability
probability
edited 30 mins ago
Key Flex
5,450828
asked 1 hour ago
Neymar
30312
edited 30 mins ago
Key Flex
5,450828
asked 1 hour ago
Neymar
30312
edited 30 mins ago
Key Flex
5,450828
edited 30 mins ago
Key Flex
5,450828
edited 30 mins ago
Key Flex
5,450828
5,450828
asked 1 hour ago
Neymar
30312
asked 1 hour ago
Neymar
30312
asked 1 hour ago
Neymar
30312
30312
$LaTeX$ Tip: use Bigr( frac 12 Bigr) to get $Bigr( frac 12 Bigr)$
– Mohammad Zuhair Khan
1 hour ago
add a comment |Â
$LaTeX$ Tip: use Bigr( frac 12 Bigr) to get $Bigr( frac 12 Bigr)$
– Mohammad Zuhair Khan
1 hour ago
$LaTeX$ Tip: use Bigr( frac 12 Bigr) to get $Bigr( frac 12 Bigr)$
– Mohammad Zuhair Khan
1 hour ago
$LaTeX$ Tip: use Bigr( frac 12 Bigr) to get $Bigr( frac 12 Bigr)$
– Mohammad Zuhair Khan
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
First, note that we have a finite number of trials, $n = 4 ($although the game goes on forever, we are only concerned with the first $4$ balls.$)$ Each trial is a Bernoulli trial - that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success $p$ is $p =dfrac12$. Since each ball is replaced after it is drawn, we have sampling with replacement, and thus independence.
Since we are dealing with a finite number of independent Bernoulli trials with a constant probability of success $p$, we use the binomial distribution
Let $X$ be the number of white balls (successes) that appear in $n = 4$ trials. Then we want to find $P(X=2)$
$$P(X=k)=dbinomnkp^k(1-p)^n-k$$
Then,
$$P(X=2)=dbinom42left(dfrac12right)^2left(1-dfrac12right)^4-2$$
$$=dbinom42left(dfrac12right)^2left(dfrac12right)^2$$
$$=dfrac38=0.375$$
answered 44 mins ago
Key Flex
5,450828
add a comment |Â
up vote
1
down vote
As far as I understand your solution, you are computing the probability that it takes $4$ draws to get $2$ white. This is not what the question asked.
You should simply be using the binomial distribution, and the answer is $frac616=frac38$.
answered 59 mins ago
David
66.5k663125
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
$checkmark !$
– David
44 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
add a comment |Â
up vote
1
down vote
$P(2W|4) = binom42cdot (frac12)^4$
$ = 6cdot frac116 = frac38$
answered 53 mins ago
Phil H
2,9742311
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First, note that we have a finite number of trials, $n = 4 ($although the game goes on forever, we are only concerned with the first $4$ balls.$)$ Each trial is a Bernoulli trial - that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success $p$ is $p =dfrac12$. Since each ball is replaced after it is drawn, we have sampling with replacement, and thus independence.
Since we are dealing with a finite number of independent Bernoulli trials with a constant probability of success $p$, we use the binomial distribution
Let $X$ be the number of white balls (successes) that appear in $n = 4$ trials. Then we want to find $P(X=2)$
$$P(X=k)=dbinomnkp^k(1-p)^n-k$$
Then,
$$P(X=2)=dbinom42left(dfrac12right)^2left(1-dfrac12right)^4-2$$
$$=dbinom42left(dfrac12right)^2left(dfrac12right)^2$$
$$=dfrac38=0.375$$
answered 44 mins ago
Key Flex
5,450828
add a comment |Â
up vote
1
down vote
accepted
First, note that we have a finite number of trials, $n = 4 ($although the game goes on forever, we are only concerned with the first $4$ balls.$)$ Each trial is a Bernoulli trial - that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success $p$ is $p =dfrac12$. Since each ball is replaced after it is drawn, we have sampling with replacement, and thus independence.
Since we are dealing with a finite number of independent Bernoulli trials with a constant probability of success $p$, we use the binomial distribution
Let $X$ be the number of white balls (successes) that appear in $n = 4$ trials. Then we want to find $P(X=2)$
$$P(X=k)=dbinomnkp^k(1-p)^n-k$$
Then,
$$P(X=2)=dbinom42left(dfrac12right)^2left(1-dfrac12right)^4-2$$
$$=dbinom42left(dfrac12right)^2left(dfrac12right)^2$$
$$=dfrac38=0.375$$
answered 44 mins ago
Key Flex
5,450828
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First, note that we have a finite number of trials, $n = 4 ($although the game goes on forever, we are only concerned with the first $4$ balls.$)$ Each trial is a Bernoulli trial - that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success $p$ is $p =dfrac12$. Since each ball is replaced after it is drawn, we have sampling with replacement, and thus independence.
Since we are dealing with a finite number of independent Bernoulli trials with a constant probability of success $p$, we use the binomial distribution
Let $X$ be the number of white balls (successes) that appear in $n = 4$ trials. Then we want to find $P(X=2)$
$$P(X=k)=dbinomnkp^k(1-p)^n-k$$
Then,
$$P(X=2)=dbinom42left(dfrac12right)^2left(1-dfrac12right)^4-2$$
$$=dbinom42left(dfrac12right)^2left(dfrac12right)^2$$
$$=dfrac38=0.375$$
answered 44 mins ago
Key Flex
5,450828
First, note that we have a finite number of trials, $n = 4 ($although the game goes on forever, we are only concerned with the first $4$ balls.$)$ Each trial is a Bernoulli trial - that is, each trial has only one of two outcomes: white and not white. Define a success as the event that a white ball is drawn. Then the probability of success $p$ is $p =dfrac12$. Since each ball is replaced after it is drawn, we have sampling with replacement, and thus independence.
Since we are dealing with a finite number of independent Bernoulli trials with a constant probability of success $p$, we use the binomial distribution
Let $X$ be the number of white balls (successes) that appear in $n = 4$ trials. Then we want to find $P(X=2)$
$$P(X=k)=dbinomnkp^k(1-p)^n-k$$
Then,
$$P(X=2)=dbinom42left(dfrac12right)^2left(1-dfrac12right)^4-2$$
$$=dbinom42left(dfrac12right)^2left(dfrac12right)^2$$
$$=dfrac38=0.375$$
answered 44 mins ago
Key Flex
5,450828
answered 44 mins ago
Key Flex
5,450828
answered 44 mins ago
Key Flex
5,450828
answered 44 mins ago
Key Flex
5,450828
5,450828
add a comment |Â
add a comment |Â
up vote
1
down vote
As far as I understand your solution, you are computing the probability that it takes $4$ draws to get $2$ white. This is not what the question asked.
You should simply be using the binomial distribution, and the answer is $frac616=frac38$.
answered 59 mins ago
David
66.5k663125
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
$checkmark !$
– David
44 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
add a comment |Â
up vote
1
down vote
As far as I understand your solution, you are computing the probability that it takes $4$ draws to get $2$ white. This is not what the question asked.
You should simply be using the binomial distribution, and the answer is $frac616=frac38$.
answered 59 mins ago
David
66.5k663125
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
$checkmark !$
– David
44 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As far as I understand your solution, you are computing the probability that it takes $4$ draws to get $2$ white. This is not what the question asked.
You should simply be using the binomial distribution, and the answer is $frac616=frac38$.
answered 59 mins ago
David
66.5k663125
As far as I understand your solution, you are computing the probability that it takes $4$ draws to get $2$ white. This is not what the question asked.
You should simply be using the binomial distribution, and the answer is $frac616=frac38$.
answered 59 mins ago
David
66.5k663125
answered 59 mins ago
David
66.5k663125
answered 59 mins ago
David
66.5k663125
answered 59 mins ago
David
66.5k663125
66.5k663125
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
$checkmark !$
– David
44 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
add a comment |Â
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
$checkmark !$
– David
44 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
Oh! So, this is bernoulli trials and we have 4 trials and we call $X$ be the number of white balls and so $P(X=2) = 4 choose 2 (1/2)^2 (1/2)^2 = 6/16 = 3/8 $. Is this correct now?
– Neymar
48 mins ago
$checkmark !$
– David
44 mins ago
$checkmark !$
– David
44 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
I dont know why I was thinking on negtive binomial. I always get confused with the wording of this problems.
– Neymar
43 mins ago
add a comment |Â
up vote
1
down vote
$P(2W|4) = binom42cdot (frac12)^4$
$ = 6cdot frac116 = frac38$
answered 53 mins ago
Phil H
2,9742311
add a comment |Â
up vote
1
down vote
$P(2W|4) = binom42cdot (frac12)^4$
$ = 6cdot frac116 = frac38$
answered 53 mins ago
Phil H
2,9742311
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$P(2W|4) = binom42cdot (frac12)^4$
$ = 6cdot frac116 = frac38$
answered 53 mins ago
Phil H
2,9742311
$P(2W|4) = binom42cdot (frac12)^4$
$ = 6cdot frac116 = frac38$
answered 53 mins ago
Phil H
2,9742311
answered 53 mins ago
Phil H
2,9742311
answered 53 mins ago
Phil H
2,9742311
answered 53 mins ago
Phil H
2,9742311
2,9742311
add a comment |Â
add a comment |Â
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$LaTeX$ Tip: use Bigr( frac 12 Bigr) to get $Bigr( frac 12 Bigr)$
– Mohammad Zuhair Khan
1 hour ago