Probability of second ball chosen is Red.
$begingroup$
Suppose, there are $5$ Red balls, $2$ Green Balls and $3$ Yellow Balls in a bag. What is the probability that the second ball taken out is a Red ball (the color of the first ball taken out can be any of the three)?
I have made the outcome tree and came up with the answer as $frac12$.
But I don't know if it is correct.
probability conditional-probability
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
$endgroup$
add a comment |
$begingroup$
Suppose, there are $5$ Red balls, $2$ Green Balls and $3$ Yellow Balls in a bag. What is the probability that the second ball taken out is a Red ball (the color of the first ball taken out can be any of the three)?
I have made the outcome tree and came up with the answer as $frac12$.
But I don't know if it is correct.
probability conditional-probability
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
$endgroup$
$begingroup$
$P=frac5{10}×frac4{9}+frac2{10}×frac5{9}+frac3{10}×frac5{9}=frac1{2}$
$endgroup$
– Thomas Shelby
Dec 28 '18 at 5:49
$begingroup$
Each of the 10 balls is equally likely to be the second one chosen, so the probability of it being red is 5/10 = 1/2 (it's simpler to ignore the color of the first ball).
$endgroup$
– Ned
Dec 28 '18 at 5:59
add a comment |
$begingroup$
Suppose, there are $5$ Red balls, $2$ Green Balls and $3$ Yellow Balls in a bag. What is the probability that the second ball taken out is a Red ball (the color of the first ball taken out can be any of the three)?
I have made the outcome tree and came up with the answer as $frac12$.
But I don't know if it is correct.
probability conditional-probability
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
$endgroup$
Suppose, there are $5$ Red balls, $2$ Green Balls and $3$ Yellow Balls in a bag. What is the probability that the second ball taken out is a Red ball (the color of the first ball taken out can be any of the three)?
I have made the outcome tree and came up with the answer as $frac12$.
But I don't know if it is correct.
probability conditional-probability
probability conditional-probability
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
edited Dec 28 '18 at 7:33
Thomas Shelby
4,1742726
4,1742726
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
asked Dec 28 '18 at 5:44
Harsh DaveHarsh Dave
31
31
$begingroup$
$P=frac5{10}×frac4{9}+frac2{10}×frac5{9}+frac3{10}×frac5{9}=frac1{2}$
$endgroup$
– Thomas Shelby
Dec 28 '18 at 5:49
$begingroup$
Each of the 10 balls is equally likely to be the second one chosen, so the probability of it being red is 5/10 = 1/2 (it's simpler to ignore the color of the first ball).
$endgroup$
– Ned
Dec 28 '18 at 5:59
add a comment |
$begingroup$
$P=frac5{10}×frac4{9}+frac2{10}×frac5{9}+frac3{10}×frac5{9}=frac1{2}$
$endgroup$
– Thomas Shelby
Dec 28 '18 at 5:49
$begingroup$
Each of the 10 balls is equally likely to be the second one chosen, so the probability of it being red is 5/10 = 1/2 (it's simpler to ignore the color of the first ball).
$endgroup$
– Ned
Dec 28 '18 at 5:59
$begingroup$
$P=frac5{10}×frac4{9}+frac2{10}×frac5{9}+frac3{10}×frac5{9}=frac1{2}$
$endgroup$
– Thomas Shelby
Dec 28 '18 at 5:49
$begingroup$
$P=frac5{10}×frac4{9}+frac2{10}×frac5{9}+frac3{10}×frac5{9}=frac1{2}$
$endgroup$
– Thomas Shelby
Dec 28 '18 at 5:49
$begingroup$
Each of the 10 balls is equally likely to be the second one chosen, so the probability of it being red is 5/10 = 1/2 (it's simpler to ignore the color of the first ball).
$endgroup$
– Ned
Dec 28 '18 at 5:59
$begingroup$
Each of the 10 balls is equally likely to be the second one chosen, so the probability of it being red is 5/10 = 1/2 (it's simpler to ignore the color of the first ball).
$endgroup$
– Ned
Dec 28 '18 at 5:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the first ball is red (which happens with probability $frac{5}{10}=frac{1}{2}$) we are left with $9$ balls of which $4$ red, so then the probability is $frac{4}{9}$ for the second one to be red.
If the first is non-red (which also happens with probability $frac{5}{10}=frac{1}{2}$) we're left with $9$ balls of which $5$ are red, so then the probability is $frac{5}{9}$ for the second one to be red.
So the "bare" or "full" probability of the second one being red is computed by conditioning on these two first outcomes and equals
$$frac{1}{2}times frac{4}{9} + frac{1}{2}times frac{5}{9}= frac{9}{18}=frac{1}{2}$$
so your answer is correct.
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the first ball is red (which happens with probability $frac{5}{10}=frac{1}{2}$) we are left with $9$ balls of which $4$ red, so then the probability is $frac{4}{9}$ for the second one to be red.
If the first is non-red (which also happens with probability $frac{5}{10}=frac{1}{2}$) we're left with $9$ balls of which $5$ are red, so then the probability is $frac{5}{9}$ for the second one to be red.
So the "bare" or "full" probability of the second one being red is computed by conditioning on these two first outcomes and equals
$$frac{1}{2}times frac{4}{9} + frac{1}{2}times frac{5}{9}= frac{9}{18}=frac{1}{2}$$
so your answer is correct.
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
$endgroup$
add a comment |
$begingroup$
If the first ball is red (which happens with probability $frac{5}{10}=frac{1}{2}$) we are left with $9$ balls of which $4$ red, so then the probability is $frac{4}{9}$ for the second one to be red.
If the first is non-red (which also happens with probability $frac{5}{10}=frac{1}{2}$) we're left with $9$ balls of which $5$ are red, so then the probability is $frac{5}{9}$ for the second one to be red.
So the "bare" or "full" probability of the second one being red is computed by conditioning on these two first outcomes and equals
$$frac{1}{2}times frac{4}{9} + frac{1}{2}times frac{5}{9}= frac{9}{18}=frac{1}{2}$$
so your answer is correct.
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
$endgroup$
add a comment |
$begingroup$
If the first ball is red (which happens with probability $frac{5}{10}=frac{1}{2}$) we are left with $9$ balls of which $4$ red, so then the probability is $frac{4}{9}$ for the second one to be red.
If the first is non-red (which also happens with probability $frac{5}{10}=frac{1}{2}$) we're left with $9$ balls of which $5$ are red, so then the probability is $frac{5}{9}$ for the second one to be red.
So the "bare" or "full" probability of the second one being red is computed by conditioning on these two first outcomes and equals
$$frac{1}{2}times frac{4}{9} + frac{1}{2}times frac{5}{9}= frac{9}{18}=frac{1}{2}$$
so your answer is correct.
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
$endgroup$
If the first ball is red (which happens with probability $frac{5}{10}=frac{1}{2}$) we are left with $9$ balls of which $4$ red, so then the probability is $frac{4}{9}$ for the second one to be red.
If the first is non-red (which also happens with probability $frac{5}{10}=frac{1}{2}$) we're left with $9$ balls of which $5$ are red, so then the probability is $frac{5}{9}$ for the second one to be red.
So the "bare" or "full" probability of the second one being red is computed by conditioning on these two first outcomes and equals
$$frac{1}{2}times frac{4}{9} + frac{1}{2}times frac{5}{9}= frac{9}{18}=frac{1}{2}$$
so your answer is correct.
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
answered Dec 28 '18 at 5:50
Henno BrandsmaHenno Brandsma
113k348121
113k348121
add a comment |
add a comment |
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$begingroup$
$P=frac5{10}×frac4{9}+frac2{10}×frac5{9}+frac3{10}×frac5{9}=frac1{2}$
$endgroup$
– Thomas Shelby
Dec 28 '18 at 5:49
$begingroup$
Each of the 10 balls is equally likely to be the second one chosen, so the probability of it being red is 5/10 = 1/2 (it's simpler to ignore the color of the first ball).
$endgroup$
– Ned
Dec 28 '18 at 5:59