Mixing
A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. Then Find the following...
$begingroup$
Question: A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. Show that the probability of at least two balls are white is $frac{67}{256}$.*
I have tried in the following way:
Number of white balls = 4
Number of black balls = 7
Number of red balls = 5
Total balls = 16
$P (text{at least $2$ balls drawn are white})$
$= P (2text{ balls drawn are white}) + P (3 text{ balls drawn are white}) +P (4 text{ balls drawn are white}) $
$=frac{^4C_2times ^{12}C_2}{^{16}C_4}+frac{^4C_3times ^{12}C_1}{^{16}C_4}+frac{^4C_4times ^{12}C_0}{^{16}C_4}$
I don't know whether my approach is correct or not.
probability combinatorics
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
$endgroup$
add a comment |
$begingroup$
Question: A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. Show that the probability of at least two balls are white is $frac{67}{256}$.*
I have tried in the following way:
Number of white balls = 4
Number of black balls = 7
Number of red balls = 5
Total balls = 16
$P (text{at least $2$ balls drawn are white})$
$= P (2text{ balls drawn are white}) + P (3 text{ balls drawn are white}) +P (4 text{ balls drawn are white}) $
$=frac{^4C_2times ^{12}C_2}{^{16}C_4}+frac{^4C_3times ^{12}C_1}{^{16}C_4}+frac{^4C_4times ^{12}C_0}{^{16}C_4}$
I don't know whether my approach is correct or not.
probability combinatorics
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
$endgroup$
$begingroup$
It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you?
$endgroup$
– The Count
Jan 18 '17 at 3:08
add a comment |
$begingroup$
Question: A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. Show that the probability of at least two balls are white is $frac{67}{256}$.*
I have tried in the following way:
Number of white balls = 4
Number of black balls = 7
Number of red balls = 5
Total balls = 16
$P (text{at least $2$ balls drawn are white})$
$= P (2text{ balls drawn are white}) + P (3 text{ balls drawn are white}) +P (4 text{ balls drawn are white}) $
$=frac{^4C_2times ^{12}C_2}{^{16}C_4}+frac{^4C_3times ^{12}C_1}{^{16}C_4}+frac{^4C_4times ^{12}C_0}{^{16}C_4}$
I don't know whether my approach is correct or not.
probability combinatorics
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
$endgroup$
Question: A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. Show that the probability of at least two balls are white is $frac{67}{256}$.*
I have tried in the following way:
Number of white balls = 4
Number of black balls = 7
Number of red balls = 5
Total balls = 16
$P (text{at least $2$ balls drawn are white})$
$= P (2text{ balls drawn are white}) + P (3 text{ balls drawn are white}) +P (4 text{ balls drawn are white}) $
$=frac{^4C_2times ^{12}C_2}{^{16}C_4}+frac{^4C_3times ^{12}C_1}{^{16}C_4}+frac{^4C_4times ^{12}C_0}{^{16}C_4}$
I don't know whether my approach is correct or not.
probability combinatorics
probability combinatorics
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
edited Jan 18 '17 at 3:11
suomynonA
5,65122557
5,65122557
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
asked Jan 18 '17 at 2:59
PrimoPrimo
1412
1412
$begingroup$
It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you?
$endgroup$
– The Count
Jan 18 '17 at 3:08
add a comment |
$begingroup$
It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you?
$endgroup$
– The Count
Jan 18 '17 at 3:08
$begingroup$
It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you?
$endgroup$
– The Count
Jan 18 '17 at 3:08
$begingroup$
It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you?
$endgroup$
– The Count
Jan 18 '17 at 3:08
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Note you're sampling without replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.
If two are white, any two of the four can be white, so there are ${4choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$
Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$
And the probability all four are white is $(4/16)^4.$
The total probability that two or more are white is the sum $$frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = frac{17152}{16^4} = frac{67}{256} $$
Your approach of using $12 choose 2$, etc is better suited to sampling without replacement.
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
add a comment |
$begingroup$
Let X be number of white balls during 4 draws, we can see X follows Binomial distribution because of drawing with replacement: Bin(4, 1/4).
So $P(X=k) = {4 choose k}frac{1}{4}^kfrac{3}{4}^{4-k} $
$ P(Xge2) = 1 - P(X=0) - P(X=1) $
Plug in 0 and 1 to the PMF and get the answer $frac{67}{256}$.
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
$endgroup$
add a comment |
$begingroup$
Well, the probability that any one of them will be white is $1/8$, correct?
So if you want at least two of them to be white, then you can find the probability that only one or zero of them will be white, then take the inverse of that.
$$begin{align*}P(text{one ball will be white}) &= frac18cdotleft(frac78right)^3cdot4=frac12cdotleft(frac78right)^3 \
P(text{no balls will be white}) &= left(frac78right)^4\
P(text{at least two will be white})&=1-P(text{one will be white}) - P(text{no balls will be white})end{align*}$$
Calculate the rest from there.
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
$endgroup$
add a comment |
$begingroup$
To follow a more systematic approach that you can use in general, start with
$$
p_{;1} = {4 over {16}}quad p_{;2} = {7 over {16}}quad p_{;3} = {5 over {16}}
$$
Because draws are with replacement, they are constant at each draw.
Now, consider that
$$
1 = p_{;1} + p_{;2} + p_{;3} = left( {p_{;1} + p_{;2} + p_{;3} } right)^{,4} = left( {p_{;1} + left( {1 - p_{;1} } right)} right)^{,4}
$$
From this, applying the binomial (or multinomial) expansion you can compute all
the concerned probabilities.
In your case
$$
Pleft( {2 le w} right) = sumlimits_{2, le ,k, le 4} {left( matrix{
4 cr
k cr} right)p_{;1} ^{,k} left( {1 - p_{;1} } right)^{,4 - k} } = {{67} over {256}}
$$
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Note you're sampling without replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.
If two are white, any two of the four can be white, so there are ${4choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$
Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$
And the probability all four are white is $(4/16)^4.$
The total probability that two or more are white is the sum $$frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = frac{17152}{16^4} = frac{67}{256} $$
Your approach of using $12 choose 2$, etc is better suited to sampling without replacement.
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
add a comment |
$begingroup$
Note you're sampling without replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.
If two are white, any two of the four can be white, so there are ${4choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$
Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$
And the probability all four are white is $(4/16)^4.$
The total probability that two or more are white is the sum $$frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = frac{17152}{16^4} = frac{67}{256} $$
Your approach of using $12 choose 2$, etc is better suited to sampling without replacement.
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
add a comment |
$begingroup$
Note you're sampling without replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.
If two are white, any two of the four can be white, so there are ${4choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$
Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$
And the probability all four are white is $(4/16)^4.$
The total probability that two or more are white is the sum $$frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = frac{17152}{16^4} = frac{67}{256} $$
Your approach of using $12 choose 2$, etc is better suited to sampling without replacement.
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
Note you're sampling without replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.
If two are white, any two of the four can be white, so there are ${4choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$
Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$
And the probability all four are white is $(4/16)^4.$
The total probability that two or more are white is the sum $$frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = frac{17152}{16^4} = frac{67}{256} $$
Your approach of using $12 choose 2$, etc is better suited to sampling without replacement.
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
answered Jan 18 '17 at 3:09
spaceisdarkgreenspaceisdarkgreen
33.4k21753
33.4k21753
add a comment |
add a comment |
$begingroup$
Let X be number of white balls during 4 draws, we can see X follows Binomial distribution because of drawing with replacement: Bin(4, 1/4).
So $P(X=k) = {4 choose k}frac{1}{4}^kfrac{3}{4}^{4-k} $
$ P(Xge2) = 1 - P(X=0) - P(X=1) $
Plug in 0 and 1 to the PMF and get the answer $frac{67}{256}$.
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
$endgroup$
add a comment |
$begingroup$
Let X be number of white balls during 4 draws, we can see X follows Binomial distribution because of drawing with replacement: Bin(4, 1/4).
So $P(X=k) = {4 choose k}frac{1}{4}^kfrac{3}{4}^{4-k} $
$ P(Xge2) = 1 - P(X=0) - P(X=1) $
Plug in 0 and 1 to the PMF and get the answer $frac{67}{256}$.
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
$endgroup$
add a comment |
$begingroup$
Let X be number of white balls during 4 draws, we can see X follows Binomial distribution because of drawing with replacement: Bin(4, 1/4).
So $P(X=k) = {4 choose k}frac{1}{4}^kfrac{3}{4}^{4-k} $
$ P(Xge2) = 1 - P(X=0) - P(X=1) $
Plug in 0 and 1 to the PMF and get the answer $frac{67}{256}$.
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
$endgroup$
Let X be number of white balls during 4 draws, we can see X follows Binomial distribution because of drawing with replacement: Bin(4, 1/4).
So $P(X=k) = {4 choose k}frac{1}{4}^kfrac{3}{4}^{4-k} $
$ P(Xge2) = 1 - P(X=0) - P(X=1) $
Plug in 0 and 1 to the PMF and get the answer $frac{67}{256}$.
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
answered Jan 18 '17 at 3:40
Ryan Y.Ryan Y.
413
413
add a comment |
add a comment |
$begingroup$
Well, the probability that any one of them will be white is $1/8$, correct?
So if you want at least two of them to be white, then you can find the probability that only one or zero of them will be white, then take the inverse of that.
$$begin{align*}P(text{one ball will be white}) &= frac18cdotleft(frac78right)^3cdot4=frac12cdotleft(frac78right)^3 \
P(text{no balls will be white}) &= left(frac78right)^4\
P(text{at least two will be white})&=1-P(text{one will be white}) - P(text{no balls will be white})end{align*}$$
Calculate the rest from there.
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
$endgroup$
add a comment |
$begingroup$
Well, the probability that any one of them will be white is $1/8$, correct?
So if you want at least two of them to be white, then you can find the probability that only one or zero of them will be white, then take the inverse of that.
$$begin{align*}P(text{one ball will be white}) &= frac18cdotleft(frac78right)^3cdot4=frac12cdotleft(frac78right)^3 \
P(text{no balls will be white}) &= left(frac78right)^4\
P(text{at least two will be white})&=1-P(text{one will be white}) - P(text{no balls will be white})end{align*}$$
Calculate the rest from there.
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
$endgroup$
add a comment |
$begingroup$
Well, the probability that any one of them will be white is $1/8$, correct?
So if you want at least two of them to be white, then you can find the probability that only one or zero of them will be white, then take the inverse of that.
$$begin{align*}P(text{one ball will be white}) &= frac18cdotleft(frac78right)^3cdot4=frac12cdotleft(frac78right)^3 \
P(text{no balls will be white}) &= left(frac78right)^4\
P(text{at least two will be white})&=1-P(text{one will be white}) - P(text{no balls will be white})end{align*}$$
Calculate the rest from there.
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
$endgroup$
Well, the probability that any one of them will be white is $1/8$, correct?
So if you want at least two of them to be white, then you can find the probability that only one or zero of them will be white, then take the inverse of that.
$$begin{align*}P(text{one ball will be white}) &= frac18cdotleft(frac78right)^3cdot4=frac12cdotleft(frac78right)^3 \
P(text{no balls will be white}) &= left(frac78right)^4\
P(text{at least two will be white})&=1-P(text{one will be white}) - P(text{no balls will be white})end{align*}$$
Calculate the rest from there.
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
answered Jan 18 '17 at 3:10
Drew ChristensenDrew Christensen
404211
404211
add a comment |
add a comment |
$begingroup$
To follow a more systematic approach that you can use in general, start with
$$
p_{;1} = {4 over {16}}quad p_{;2} = {7 over {16}}quad p_{;3} = {5 over {16}}
$$
Because draws are with replacement, they are constant at each draw.
Now, consider that
$$
1 = p_{;1} + p_{;2} + p_{;3} = left( {p_{;1} + p_{;2} + p_{;3} } right)^{,4} = left( {p_{;1} + left( {1 - p_{;1} } right)} right)^{,4}
$$
From this, applying the binomial (or multinomial) expansion you can compute all
the concerned probabilities.
In your case
$$
Pleft( {2 le w} right) = sumlimits_{2, le ,k, le 4} {left( matrix{
4 cr
k cr} right)p_{;1} ^{,k} left( {1 - p_{;1} } right)^{,4 - k} } = {{67} over {256}}
$$
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
To follow a more systematic approach that you can use in general, start with
$$
p_{;1} = {4 over {16}}quad p_{;2} = {7 over {16}}quad p_{;3} = {5 over {16}}
$$
Because draws are with replacement, they are constant at each draw.
Now, consider that
$$
1 = p_{;1} + p_{;2} + p_{;3} = left( {p_{;1} + p_{;2} + p_{;3} } right)^{,4} = left( {p_{;1} + left( {1 - p_{;1} } right)} right)^{,4}
$$
From this, applying the binomial (or multinomial) expansion you can compute all
the concerned probabilities.
In your case
$$
Pleft( {2 le w} right) = sumlimits_{2, le ,k, le 4} {left( matrix{
4 cr
k cr} right)p_{;1} ^{,k} left( {1 - p_{;1} } right)^{,4 - k} } = {{67} over {256}}
$$
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
To follow a more systematic approach that you can use in general, start with
$$
p_{;1} = {4 over {16}}quad p_{;2} = {7 over {16}}quad p_{;3} = {5 over {16}}
$$
Because draws are with replacement, they are constant at each draw.
Now, consider that
$$
1 = p_{;1} + p_{;2} + p_{;3} = left( {p_{;1} + p_{;2} + p_{;3} } right)^{,4} = left( {p_{;1} + left( {1 - p_{;1} } right)} right)^{,4}
$$
From this, applying the binomial (or multinomial) expansion you can compute all
the concerned probabilities.
In your case
$$
Pleft( {2 le w} right) = sumlimits_{2, le ,k, le 4} {left( matrix{
4 cr
k cr} right)p_{;1} ^{,k} left( {1 - p_{;1} } right)^{,4 - k} } = {{67} over {256}}
$$
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
$endgroup$
To follow a more systematic approach that you can use in general, start with
$$
p_{;1} = {4 over {16}}quad p_{;2} = {7 over {16}}quad p_{;3} = {5 over {16}}
$$
Because draws are with replacement, they are constant at each draw.
Now, consider that
$$
1 = p_{;1} + p_{;2} + p_{;3} = left( {p_{;1} + p_{;2} + p_{;3} } right)^{,4} = left( {p_{;1} + left( {1 - p_{;1} } right)} right)^{,4}
$$
From this, applying the binomial (or multinomial) expansion you can compute all
the concerned probabilities.
In your case
$$
Pleft( {2 le w} right) = sumlimits_{2, le ,k, le 4} {left( matrix{
4 cr
k cr} right)p_{;1} ^{,k} left( {1 - p_{;1} } right)^{,4 - k} } = {{67} over {256}}
$$
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
answered Jan 22 at 19:03
G CabG Cab
19.9k31340
19.9k31340
add a comment |
add a comment |
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$begingroup$
It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you?
$endgroup$
– The Count
Jan 18 '17 at 3:08