Mixing
Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls
$begingroup$
Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls. What is the probability of getting a blue ball if
(a) You select a ball at random from the first urn?
(b) You select an urn at random and then select a ball from it at random?
(c) You discard two balls from the second urn and select the last ball?
MY ATTEMPT
Let us denote by $B$ the event "the user has selected a blue ball". There are three possible configurations for the first urn: there is no blue ball, there is one blue ball or there are two blue balls.
Let us denote by $B_{k}$ the event "there is $k$ blue balls in the first urn" where $0leq k leq 2$, from whence we obtain the partition ${B_{0},B_{1},B_{2}}$. Precisely speaking, it results that
begin{align*}
P(B) & = P(Bcap(B_{0}cup B_{1}cup B_{2})) = P(Bcap B_{0}) + P(Bcap B_{1}) + P(Bcap B_{2}) =\\
& = 0 + P(B|B_{1})P(B_{1}) + P(B|B_{2})P(B_{2}) = frac{1}{3}times P(B_{1}) + frac{2}{3}times P(B_{2})
end{align*}
Here is my problem: how do we calculate $P(B_{1})$ and $P(B_{2})$?
b) It suffices to notice that the selection of the urn and the selection of the ball are independent.
c) I do not know how to solve.
Am I on the right track? Any help is appreciated. Thanks in advance.
probability probability-theory conditional-probability
asked Jan 11 at 1:18
user1337user1337
48410
$endgroup$
add a comment |
$begingroup$
Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls. What is the probability of getting a blue ball if
(a) You select a ball at random from the first urn?
(b) You select an urn at random and then select a ball from it at random?
(c) You discard two balls from the second urn and select the last ball?
MY ATTEMPT
Let us denote by $B$ the event "the user has selected a blue ball". There are three possible configurations for the first urn: there is no blue ball, there is one blue ball or there are two blue balls.
Let us denote by $B_{k}$ the event "there is $k$ blue balls in the first urn" where $0leq k leq 2$, from whence we obtain the partition ${B_{0},B_{1},B_{2}}$. Precisely speaking, it results that
begin{align*}
P(B) & = P(Bcap(B_{0}cup B_{1}cup B_{2})) = P(Bcap B_{0}) + P(Bcap B_{1}) + P(Bcap B_{2}) =\\
& = 0 + P(B|B_{1})P(B_{1}) + P(B|B_{2})P(B_{2}) = frac{1}{3}times P(B_{1}) + frac{2}{3}times P(B_{2})
end{align*}
Here is my problem: how do we calculate $P(B_{1})$ and $P(B_{2})$?
b) It suffices to notice that the selection of the urn and the selection of the ball are independent.
c) I do not know how to solve.
Am I on the right track? Any help is appreciated. Thanks in advance.
probability probability-theory conditional-probability
asked Jan 11 at 1:18
user1337user1337
48410
$endgroup$
add a comment |
$begingroup$
Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls. What is the probability of getting a blue ball if
(a) You select a ball at random from the first urn?
(b) You select an urn at random and then select a ball from it at random?
(c) You discard two balls from the second urn and select the last ball?
MY ATTEMPT
Let us denote by $B$ the event "the user has selected a blue ball". There are three possible configurations for the first urn: there is no blue ball, there is one blue ball or there are two blue balls.
Let us denote by $B_{k}$ the event "there is $k$ blue balls in the first urn" where $0leq k leq 2$, from whence we obtain the partition ${B_{0},B_{1},B_{2}}$. Precisely speaking, it results that
begin{align*}
P(B) & = P(Bcap(B_{0}cup B_{1}cup B_{2})) = P(Bcap B_{0}) + P(Bcap B_{1}) + P(Bcap B_{2}) =\\
& = 0 + P(B|B_{1})P(B_{1}) + P(B|B_{2})P(B_{2}) = frac{1}{3}times P(B_{1}) + frac{2}{3}times P(B_{2})
end{align*}
Here is my problem: how do we calculate $P(B_{1})$ and $P(B_{2})$?
b) It suffices to notice that the selection of the urn and the selection of the ball are independent.
c) I do not know how to solve.
Am I on the right track? Any help is appreciated. Thanks in advance.
probability probability-theory conditional-probability
asked Jan 11 at 1:18
user1337user1337
48410
$endgroup$
Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls. What is the probability of getting a blue ball if
(a) You select a ball at random from the first urn?
(b) You select an urn at random and then select a ball from it at random?
(c) You discard two balls from the second urn and select the last ball?
MY ATTEMPT
Let us denote by $B$ the event "the user has selected a blue ball". There are three possible configurations for the first urn: there is no blue ball, there is one blue ball or there are two blue balls.
Let us denote by $B_{k}$ the event "there is $k$ blue balls in the first urn" where $0leq k leq 2$, from whence we obtain the partition ${B_{0},B_{1},B_{2}}$. Precisely speaking, it results that
begin{align*}
P(B) & = P(Bcap(B_{0}cup B_{1}cup B_{2})) = P(Bcap B_{0}) + P(Bcap B_{1}) + P(Bcap B_{2}) =\\
& = 0 + P(B|B_{1})P(B_{1}) + P(B|B_{2})P(B_{2}) = frac{1}{3}times P(B_{1}) + frac{2}{3}times P(B_{2})
end{align*}
Here is my problem: how do we calculate $P(B_{1})$ and $P(B_{2})$?
b) It suffices to notice that the selection of the urn and the selection of the ball are independent.
c) I do not know how to solve.
Am I on the right track? Any help is appreciated. Thanks in advance.
probability probability-theory conditional-probability
probability probability-theory conditional-probability
asked Jan 11 at 1:18
user1337user1337
48410
asked Jan 11 at 1:18
user1337user1337
48410
edited Jan 11 at 1:31
user1337
asked Jan 11 at 1:18
user1337user1337
48410
asked Jan 11 at 1:18
user1337user1337
48410
asked Jan 11 at 1:18
user1337user1337
48410
48410
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2 Answers
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$begingroup$
Fundamentally, the answer to all three parts is $frac 13$ by symmetry. There is nothing to distinguish the ball you have picked.
You could note that $P(B_0)=P(B_2)$ because if one urn has two blues the second has none. To get $P(B_0)$ there are $20$ ways to choose three balls of six, and four of them have three red balls. So $P(B_0)=P(B_2)=frac 15$ and $P(B_1)=frac 35$. Then $frac 13cdot frac 35+frac 23 cdot frac 15=frac 13$ as promised.
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
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$begingroup$
You have ${6choose3}=20$ ways to choose the $3$ balls in the first urn. There are $4$ ways to choose $2$ blue balls and $1$ red ball, $12$ ways to choose $1$ blue ball and $2$ red balls, and $4$ ways to choose $3$ red balls. In your notation, $P(B_1)={12over20}={3over5},$ and $P(B_2)={4over20}={1over5}.$
The answer to part b) is the same as the answer to part a) since the distribution of balls is the same in the two urns. The answer to part c) is the same as the answer to the other two parts. Discarding the first two balls and picking the last is just another way of selecting a ball at random. You are equally likely to choose any of the three balls. So, part c) is just picking a ball at random from the second urn.
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
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2 Answers
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$begingroup$
Fundamentally, the answer to all three parts is $frac 13$ by symmetry. There is nothing to distinguish the ball you have picked.
You could note that $P(B_0)=P(B_2)$ because if one urn has two blues the second has none. To get $P(B_0)$ there are $20$ ways to choose three balls of six, and four of them have three red balls. So $P(B_0)=P(B_2)=frac 15$ and $P(B_1)=frac 35$. Then $frac 13cdot frac 35+frac 23 cdot frac 15=frac 13$ as promised.
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
Fundamentally, the answer to all three parts is $frac 13$ by symmetry. There is nothing to distinguish the ball you have picked.
You could note that $P(B_0)=P(B_2)$ because if one urn has two blues the second has none. To get $P(B_0)$ there are $20$ ways to choose three balls of six, and four of them have three red balls. So $P(B_0)=P(B_2)=frac 15$ and $P(B_1)=frac 35$. Then $frac 13cdot frac 35+frac 23 cdot frac 15=frac 13$ as promised.
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
Fundamentally, the answer to all three parts is $frac 13$ by symmetry. There is nothing to distinguish the ball you have picked.
You could note that $P(B_0)=P(B_2)$ because if one urn has two blues the second has none. To get $P(B_0)$ there are $20$ ways to choose three balls of six, and four of them have three red balls. So $P(B_0)=P(B_2)=frac 15$ and $P(B_1)=frac 35$. Then $frac 13cdot frac 35+frac 23 cdot frac 15=frac 13$ as promised.
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
Fundamentally, the answer to all three parts is $frac 13$ by symmetry. There is nothing to distinguish the ball you have picked.
You could note that $P(B_0)=P(B_2)$ because if one urn has two blues the second has none. To get $P(B_0)$ there are $20$ ways to choose three balls of six, and four of them have three red balls. So $P(B_0)=P(B_2)=frac 15$ and $P(B_1)=frac 35$. Then $frac 13cdot frac 35+frac 23 cdot frac 15=frac 13$ as promised.
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
answered Jan 11 at 1:39
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
$begingroup$
You have ${6choose3}=20$ ways to choose the $3$ balls in the first urn. There are $4$ ways to choose $2$ blue balls and $1$ red ball, $12$ ways to choose $1$ blue ball and $2$ red balls, and $4$ ways to choose $3$ red balls. In your notation, $P(B_1)={12over20}={3over5},$ and $P(B_2)={4over20}={1over5}.$
The answer to part b) is the same as the answer to part a) since the distribution of balls is the same in the two urns. The answer to part c) is the same as the answer to the other two parts. Discarding the first two balls and picking the last is just another way of selecting a ball at random. You are equally likely to choose any of the three balls. So, part c) is just picking a ball at random from the second urn.
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
$endgroup$
add a comment |
$begingroup$
You have ${6choose3}=20$ ways to choose the $3$ balls in the first urn. There are $4$ ways to choose $2$ blue balls and $1$ red ball, $12$ ways to choose $1$ blue ball and $2$ red balls, and $4$ ways to choose $3$ red balls. In your notation, $P(B_1)={12over20}={3over5},$ and $P(B_2)={4over20}={1over5}.$
The answer to part b) is the same as the answer to part a) since the distribution of balls is the same in the two urns. The answer to part c) is the same as the answer to the other two parts. Discarding the first two balls and picking the last is just another way of selecting a ball at random. You are equally likely to choose any of the three balls. So, part c) is just picking a ball at random from the second urn.
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
$endgroup$
add a comment |
$begingroup$
You have ${6choose3}=20$ ways to choose the $3$ balls in the first urn. There are $4$ ways to choose $2$ blue balls and $1$ red ball, $12$ ways to choose $1$ blue ball and $2$ red balls, and $4$ ways to choose $3$ red balls. In your notation, $P(B_1)={12over20}={3over5},$ and $P(B_2)={4over20}={1over5}.$
The answer to part b) is the same as the answer to part a) since the distribution of balls is the same in the two urns. The answer to part c) is the same as the answer to the other two parts. Discarding the first two balls and picking the last is just another way of selecting a ball at random. You are equally likely to choose any of the three balls. So, part c) is just picking a ball at random from the second urn.
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
$endgroup$
You have ${6choose3}=20$ ways to choose the $3$ balls in the first urn. There are $4$ ways to choose $2$ blue balls and $1$ red ball, $12$ ways to choose $1$ blue ball and $2$ red balls, and $4$ ways to choose $3$ red balls. In your notation, $P(B_1)={12over20}={3over5},$ and $P(B_2)={4over20}={1over5}.$
The answer to part b) is the same as the answer to part a) since the distribution of balls is the same in the two urns. The answer to part c) is the same as the answer to the other two parts. Discarding the first two balls and picking the last is just another way of selecting a ball at random. You are equally likely to choose any of the three balls. So, part c) is just picking a ball at random from the second urn.
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
answered Jan 11 at 1:35
saulspatzsaulspatz
14.8k21329
14.8k21329
add a comment |
add a comment |
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