Mixing
Write down the sample space of this experiment and give the distribution.
$begingroup$
An urn contains $4$ red balls and $6$ black balls. You draw with replacement $2$ balls.
a) Write down the sample space of this experiment and list $3$ events $A$, $B$ and $C$, whose intersection is not empty (i.e. $A ∩ B ∩ C= ∅$).
b) Let $X$ be the random variable that counts the number of red balls among those extracted. What is its distribution?
For a) The sample space is $Omega={W,R}$ , so white and red (the set whose elements describe the outcomes of the experiment), correct?
For the events in order to have intersection what should I consider? If I choose A={W,W}, B={R,R}, C={R,W} I think that should be an option, what do you think?
For b) I think that the distribution is a Binomial Distribution, it's correct?
probability polya-urn-model
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
$endgroup$
add a comment |
$begingroup$
An urn contains $4$ red balls and $6$ black balls. You draw with replacement $2$ balls.
a) Write down the sample space of this experiment and list $3$ events $A$, $B$ and $C$, whose intersection is not empty (i.e. $A ∩ B ∩ C= ∅$).
b) Let $X$ be the random variable that counts the number of red balls among those extracted. What is its distribution?
For a) The sample space is $Omega={W,R}$ , so white and red (the set whose elements describe the outcomes of the experiment), correct?
For the events in order to have intersection what should I consider? If I choose A={W,W}, B={R,R}, C={R,W} I think that should be an option, what do you think?
For b) I think that the distribution is a Binomial Distribution, it's correct?
probability polya-urn-model
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
$endgroup$
add a comment |
$begingroup$
An urn contains $4$ red balls and $6$ black balls. You draw with replacement $2$ balls.
a) Write down the sample space of this experiment and list $3$ events $A$, $B$ and $C$, whose intersection is not empty (i.e. $A ∩ B ∩ C= ∅$).
b) Let $X$ be the random variable that counts the number of red balls among those extracted. What is its distribution?
For a) The sample space is $Omega={W,R}$ , so white and red (the set whose elements describe the outcomes of the experiment), correct?
For the events in order to have intersection what should I consider? If I choose A={W,W}, B={R,R}, C={R,W} I think that should be an option, what do you think?
For b) I think that the distribution is a Binomial Distribution, it's correct?
probability polya-urn-model
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
$endgroup$
An urn contains $4$ red balls and $6$ black balls. You draw with replacement $2$ balls.
a) Write down the sample space of this experiment and list $3$ events $A$, $B$ and $C$, whose intersection is not empty (i.e. $A ∩ B ∩ C= ∅$).
b) Let $X$ be the random variable that counts the number of red balls among those extracted. What is its distribution?
For a) The sample space is $Omega={W,R}$ , so white and red (the set whose elements describe the outcomes of the experiment), correct?
For the events in order to have intersection what should I consider? If I choose A={W,W}, B={R,R}, C={R,W} I think that should be an option, what do you think?
For b) I think that the distribution is a Binomial Distribution, it's correct?
probability polya-urn-model
probability polya-urn-model
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
asked Jan 19 at 21:16
Luke MarciLuke Marci
856
856
add a comment |
add a comment |
1 Answer
1
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$begingroup$
For (a), $Omega={B,R}^2={BB,BR,RB,RR}$. An example of 3 events whose intersection is not empty would be
$$
A={BB}, B={BB, BR, RB}, C={BB,RR}.
$$
$A=$ "both balls are black", $B=$ "at least one ball is black", and $C=$ "both balls are of the same color".
For (b), $mathsf{P}(X=0)=mathsf{P}({BB})=(6/10)^2$, $mathsf{P}(X=1)=mathsf{P}({BR,RB})=2times 24/100$, and so on. In general,
$$
mathsf{P}(X=k)=binom{2}{k}left(frac{4}{10}right)^kleft(frac{6}{10}right)^{2-k}.
$$
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
$endgroup$
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For (a), $Omega={B,R}^2={BB,BR,RB,RR}$. An example of 3 events whose intersection is not empty would be
$$
A={BB}, B={BB, BR, RB}, C={BB,RR}.
$$
$A=$ "both balls are black", $B=$ "at least one ball is black", and $C=$ "both balls are of the same color".
For (b), $mathsf{P}(X=0)=mathsf{P}({BB})=(6/10)^2$, $mathsf{P}(X=1)=mathsf{P}({BR,RB})=2times 24/100$, and so on. In general,
$$
mathsf{P}(X=k)=binom{2}{k}left(frac{4}{10}right)^kleft(frac{6}{10}right)^{2-k}.
$$
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
$endgroup$
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
add a comment |
$begingroup$
For (a), $Omega={B,R}^2={BB,BR,RB,RR}$. An example of 3 events whose intersection is not empty would be
$$
A={BB}, B={BB, BR, RB}, C={BB,RR}.
$$
$A=$ "both balls are black", $B=$ "at least one ball is black", and $C=$ "both balls are of the same color".
For (b), $mathsf{P}(X=0)=mathsf{P}({BB})=(6/10)^2$, $mathsf{P}(X=1)=mathsf{P}({BR,RB})=2times 24/100$, and so on. In general,
$$
mathsf{P}(X=k)=binom{2}{k}left(frac{4}{10}right)^kleft(frac{6}{10}right)^{2-k}.
$$
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
$endgroup$
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
add a comment |
$begingroup$
For (a), $Omega={B,R}^2={BB,BR,RB,RR}$. An example of 3 events whose intersection is not empty would be
$$
A={BB}, B={BB, BR, RB}, C={BB,RR}.
$$
$A=$ "both balls are black", $B=$ "at least one ball is black", and $C=$ "both balls are of the same color".
For (b), $mathsf{P}(X=0)=mathsf{P}({BB})=(6/10)^2$, $mathsf{P}(X=1)=mathsf{P}({BR,RB})=2times 24/100$, and so on. In general,
$$
mathsf{P}(X=k)=binom{2}{k}left(frac{4}{10}right)^kleft(frac{6}{10}right)^{2-k}.
$$
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
$endgroup$
For (a), $Omega={B,R}^2={BB,BR,RB,RR}$. An example of 3 events whose intersection is not empty would be
$$
A={BB}, B={BB, BR, RB}, C={BB,RR}.
$$
$A=$ "both balls are black", $B=$ "at least one ball is black", and $C=$ "both balls are of the same color".
For (b), $mathsf{P}(X=0)=mathsf{P}({BB})=(6/10)^2$, $mathsf{P}(X=1)=mathsf{P}({BR,RB})=2times 24/100$, and so on. In general,
$$
mathsf{P}(X=k)=binom{2}{k}left(frac{4}{10}right)^kleft(frac{6}{10}right)^{2-k}.
$$
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
edited Jan 19 at 22:38
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
answered Jan 19 at 22:31
d.k.o.d.k.o.
10k629
10k629
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
add a comment |
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
Thanks for the answer, how did you calculate 4/10? It is the probability that both ball are black, right? shouldn't be 1/4? :)
$endgroup$
– Luke Marci
Jan 20 at 20:44
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
$4/10$ is the probability of drawing a red ball in a single trial.
$endgroup$
– d.k.o.
Jan 20 at 20:48
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
$begingroup$
you are right, thanks again!
$endgroup$
– Luke Marci
Jan 20 at 20:49
add a comment |
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