Mixing
drawing 2 red balls from box contains 3 red balls and 7 white balls
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I have tried to find a solution for this problem a lot but I couldn't solve it.
Consider a box containing 3 red ball and 7 white balls. Suppose that balls are drawn one at a time, at random, without replacement from this box until two red balls are obtained. Let X denote the number of the draw on which the second red ball is obtained. Find the probability mass function of X
all what I could do is to get the range for X.
where X = 2 , 3, 4,5,6,7,8,9
probability combinations
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
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|
show 2 more comments
$begingroup$
I have tried to find a solution for this problem a lot but I couldn't solve it.
Consider a box containing 3 red ball and 7 white balls. Suppose that balls are drawn one at a time, at random, without replacement from this box until two red balls are obtained. Let X denote the number of the draw on which the second red ball is obtained. Find the probability mass function of X
all what I could do is to get the range for X.
where X = 2 , 3, 4,5,6,7,8,9
probability combinations
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
$endgroup$
$begingroup$
Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=frac 3{10}times frac 2{9}$ for example. Now do the rest.
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– lulu
Jan 20 at 17:14
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the problem here is i can't know how many times i will draw balls so i can not decide the probability very well.
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:22
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maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:23
$begingroup$
It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance?
$endgroup$
– lulu
Jan 20 at 17:33
$begingroup$
okay I think it will be (((3C1)*(7C1))/(10C2))*2/8
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:58
|
show 2 more comments
$begingroup$
I have tried to find a solution for this problem a lot but I couldn't solve it.
Consider a box containing 3 red ball and 7 white balls. Suppose that balls are drawn one at a time, at random, without replacement from this box until two red balls are obtained. Let X denote the number of the draw on which the second red ball is obtained. Find the probability mass function of X
all what I could do is to get the range for X.
where X = 2 , 3, 4,5,6,7,8,9
probability combinations
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
$endgroup$
I have tried to find a solution for this problem a lot but I couldn't solve it.
Consider a box containing 3 red ball and 7 white balls. Suppose that balls are drawn one at a time, at random, without replacement from this box until two red balls are obtained. Let X denote the number of the draw on which the second red ball is obtained. Find the probability mass function of X
all what I could do is to get the range for X.
where X = 2 , 3, 4,5,6,7,8,9
probability combinations
probability combinations
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
asked Jan 20 at 17:10
Mo'men MustafaMo'men Mustafa
54
54
$begingroup$
Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=frac 3{10}times frac 2{9}$ for example. Now do the rest.
$endgroup$
– lulu
Jan 20 at 17:14
$begingroup$
the problem here is i can't know how many times i will draw balls so i can not decide the probability very well.
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:22
$begingroup$
maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:23
$begingroup$
It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance?
$endgroup$
– lulu
Jan 20 at 17:33
$begingroup$
okay I think it will be (((3C1)*(7C1))/(10C2))*2/8
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:58
|
show 2 more comments
$begingroup$
Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=frac 3{10}times frac 2{9}$ for example. Now do the rest.
$endgroup$
– lulu
Jan 20 at 17:14
$begingroup$
the problem here is i can't know how many times i will draw balls so i can not decide the probability very well.
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:22
$begingroup$
maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:23
$begingroup$
It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance?
$endgroup$
– lulu
Jan 20 at 17:33
$begingroup$
okay I think it will be (((3C1)*(7C1))/(10C2))*2/8
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:58
$begingroup$
Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=frac 3{10}times frac 2{9}$ for example. Now do the rest.
$endgroup$
– lulu
Jan 20 at 17:14
$begingroup$
Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=frac 3{10}times frac 2{9}$ for example. Now do the rest.
$endgroup$
– lulu
Jan 20 at 17:14
$begingroup$
the problem here is i can't know how many times i will draw balls so i can not decide the probability very well.
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:22
$begingroup$
the problem here is i can't know how many times i will draw balls so i can not decide the probability very well.
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:22
$begingroup$
maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:23
$begingroup$
maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:23
$begingroup$
It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance?
$endgroup$
– lulu
Jan 20 at 17:33
$begingroup$
It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance?
$endgroup$
– lulu
Jan 20 at 17:33
$begingroup$
okay I think it will be (((3C1)*(7C1))/(10C2))*2/8
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:58
$begingroup$
okay I think it will be (((3C1)*(7C1))/(10C2))*2/8
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:58
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let's compute $P(X=i)$.
To be in that scenario, the first $i-1$ choices must consist of exactly $1$ red and $i-2$ whites. There are $binom 7{i-2}$ ways to choose the whites and $3$ ways to choose the reds. Thus there are $3times binom 7{i-2}$ ways to choose the first $i-1$. Of course there are $binom {10}{i-1}$ ways to do it with out restriction. Thus the probability that the first $i-1$ contain exactly $1$ red is $$3times binom 7{i-2}Big /binom {10}{i-1}$$
Having successfully chosen the first $i-1$ we now require that the $i^{th}$ choice be red. That has probability $frac 2{10-(i-1)}=frac 2{11-i}$. Thus the answer is $$left(3times binom 7{i-2}Big /binom {10}{i-1}right)times frac 2{11-i}$$
answered Jan 20 at 18:02
lulululu
42.8k25080
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let's compute $P(X=i)$.
To be in that scenario, the first $i-1$ choices must consist of exactly $1$ red and $i-2$ whites. There are $binom 7{i-2}$ ways to choose the whites and $3$ ways to choose the reds. Thus there are $3times binom 7{i-2}$ ways to choose the first $i-1$. Of course there are $binom {10}{i-1}$ ways to do it with out restriction. Thus the probability that the first $i-1$ contain exactly $1$ red is $$3times binom 7{i-2}Big /binom {10}{i-1}$$
Having successfully chosen the first $i-1$ we now require that the $i^{th}$ choice be red. That has probability $frac 2{10-(i-1)}=frac 2{11-i}$. Thus the answer is $$left(3times binom 7{i-2}Big /binom {10}{i-1}right)times frac 2{11-i}$$
answered Jan 20 at 18:02
lulululu
42.8k25080
$endgroup$
add a comment |
$begingroup$
Let's compute $P(X=i)$.
To be in that scenario, the first $i-1$ choices must consist of exactly $1$ red and $i-2$ whites. There are $binom 7{i-2}$ ways to choose the whites and $3$ ways to choose the reds. Thus there are $3times binom 7{i-2}$ ways to choose the first $i-1$. Of course there are $binom {10}{i-1}$ ways to do it with out restriction. Thus the probability that the first $i-1$ contain exactly $1$ red is $$3times binom 7{i-2}Big /binom {10}{i-1}$$
Having successfully chosen the first $i-1$ we now require that the $i^{th}$ choice be red. That has probability $frac 2{10-(i-1)}=frac 2{11-i}$. Thus the answer is $$left(3times binom 7{i-2}Big /binom {10}{i-1}right)times frac 2{11-i}$$
answered Jan 20 at 18:02
lulululu
42.8k25080
$endgroup$
add a comment |
$begingroup$
Let's compute $P(X=i)$.
To be in that scenario, the first $i-1$ choices must consist of exactly $1$ red and $i-2$ whites. There are $binom 7{i-2}$ ways to choose the whites and $3$ ways to choose the reds. Thus there are $3times binom 7{i-2}$ ways to choose the first $i-1$. Of course there are $binom {10}{i-1}$ ways to do it with out restriction. Thus the probability that the first $i-1$ contain exactly $1$ red is $$3times binom 7{i-2}Big /binom {10}{i-1}$$
Having successfully chosen the first $i-1$ we now require that the $i^{th}$ choice be red. That has probability $frac 2{10-(i-1)}=frac 2{11-i}$. Thus the answer is $$left(3times binom 7{i-2}Big /binom {10}{i-1}right)times frac 2{11-i}$$
answered Jan 20 at 18:02
lulululu
42.8k25080
$endgroup$
Let's compute $P(X=i)$.
To be in that scenario, the first $i-1$ choices must consist of exactly $1$ red and $i-2$ whites. There are $binom 7{i-2}$ ways to choose the whites and $3$ ways to choose the reds. Thus there are $3times binom 7{i-2}$ ways to choose the first $i-1$. Of course there are $binom {10}{i-1}$ ways to do it with out restriction. Thus the probability that the first $i-1$ contain exactly $1$ red is $$3times binom 7{i-2}Big /binom {10}{i-1}$$
Having successfully chosen the first $i-1$ we now require that the $i^{th}$ choice be red. That has probability $frac 2{10-(i-1)}=frac 2{11-i}$. Thus the answer is $$left(3times binom 7{i-2}Big /binom {10}{i-1}right)times frac 2{11-i}$$
answered Jan 20 at 18:02
lulululu
42.8k25080
answered Jan 20 at 18:02
lulululu
42.8k25080
answered Jan 20 at 18:02
lulululu
42.8k25080
answered Jan 20 at 18:02
lulululu
42.8k25080
42.8k25080
add a comment |
add a comment |
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$begingroup$
Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=frac 3{10}times frac 2{9}$ for example. Now do the rest.
$endgroup$
– lulu
Jan 20 at 17:14
$begingroup$
the problem here is i can't know how many times i will draw balls so i can not decide the probability very well.
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:22
$begingroup$
maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:23
$begingroup$
It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance?
$endgroup$
– lulu
Jan 20 at 17:33
$begingroup$
okay I think it will be (((3C1)*(7C1))/(10C2))*2/8
$endgroup$
– Mo'men Mustafa
Jan 20 at 17:58