One urn with white balls, another with 10 black and 30 white. Probability of drawing black after seeing...
$begingroup$
This is a problem from Y.A. Rozanov book "Probability theory: A concise course", and seems to be simple, yet the answer given disagrees with what I arrived at. Text verbatim (ex. 15) is following
Exercise 15/p36
One urn contains only white balls, while another urn contains 30 white
and 10 black balls. An urn is selected at random and then a ball is
drawn (at random) from the urn. The ball turns out to be white and is
then put back into the urn. What is the probability that another ball
drawn from the same urn will be black?
The answer given is $P = frac{3}{28}$
If the above is correct, my solution must be flawed somehow:
Let $P(x_k)$ be the probability that ball $x$ is drawn in $k-th$ draw.
Then we seek the probability $P(b_2|w_1)$. Given that urns were selected at random, and we don't which one we ended up with
$$P(b_2|w_1) = P(b_2|w_1|u_1)P(u_1) + P(b_2|w_1|u_2) P(u_2)$$
where $P(u_i) = frac{1}{2}$ is the probability that $i-th$ urn was selected. The first of these terms vanishes, as urn 1 does not contain any black balls, hence the probability of seeing a black ball is 0.
For the second, as the white ball is put back into the urn, the second draw is obviously independent with the first, hence
$$P(b_2|w_1|u_2) = frac{1}{4}$$
Thus the answer would come to be
$$P(b_2|w_1) = frac{1}{8}$$
which is obviously very different from given answer.
Where is the mistake?
probability probability-theory
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
$endgroup$
add a comment |
$begingroup$
This is a problem from Y.A. Rozanov book "Probability theory: A concise course", and seems to be simple, yet the answer given disagrees with what I arrived at. Text verbatim (ex. 15) is following
Exercise 15/p36
One urn contains only white balls, while another urn contains 30 white
and 10 black balls. An urn is selected at random and then a ball is
drawn (at random) from the urn. The ball turns out to be white and is
then put back into the urn. What is the probability that another ball
drawn from the same urn will be black?
The answer given is $P = frac{3}{28}$
If the above is correct, my solution must be flawed somehow:
Let $P(x_k)$ be the probability that ball $x$ is drawn in $k-th$ draw.
Then we seek the probability $P(b_2|w_1)$. Given that urns were selected at random, and we don't which one we ended up with
$$P(b_2|w_1) = P(b_2|w_1|u_1)P(u_1) + P(b_2|w_1|u_2) P(u_2)$$
where $P(u_i) = frac{1}{2}$ is the probability that $i-th$ urn was selected. The first of these terms vanishes, as urn 1 does not contain any black balls, hence the probability of seeing a black ball is 0.
For the second, as the white ball is put back into the urn, the second draw is obviously independent with the first, hence
$$P(b_2|w_1|u_2) = frac{1}{4}$$
Thus the answer would come to be
$$P(b_2|w_1) = frac{1}{8}$$
which is obviously very different from given answer.
Where is the mistake?
probability probability-theory
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
$endgroup$
add a comment |
$begingroup$
This is a problem from Y.A. Rozanov book "Probability theory: A concise course", and seems to be simple, yet the answer given disagrees with what I arrived at. Text verbatim (ex. 15) is following
Exercise 15/p36
One urn contains only white balls, while another urn contains 30 white
and 10 black balls. An urn is selected at random and then a ball is
drawn (at random) from the urn. The ball turns out to be white and is
then put back into the urn. What is the probability that another ball
drawn from the same urn will be black?
The answer given is $P = frac{3}{28}$
If the above is correct, my solution must be flawed somehow:
Let $P(x_k)$ be the probability that ball $x$ is drawn in $k-th$ draw.
Then we seek the probability $P(b_2|w_1)$. Given that urns were selected at random, and we don't which one we ended up with
$$P(b_2|w_1) = P(b_2|w_1|u_1)P(u_1) + P(b_2|w_1|u_2) P(u_2)$$
where $P(u_i) = frac{1}{2}$ is the probability that $i-th$ urn was selected. The first of these terms vanishes, as urn 1 does not contain any black balls, hence the probability of seeing a black ball is 0.
For the second, as the white ball is put back into the urn, the second draw is obviously independent with the first, hence
$$P(b_2|w_1|u_2) = frac{1}{4}$$
Thus the answer would come to be
$$P(b_2|w_1) = frac{1}{8}$$
which is obviously very different from given answer.
Where is the mistake?
probability probability-theory
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
$endgroup$
This is a problem from Y.A. Rozanov book "Probability theory: A concise course", and seems to be simple, yet the answer given disagrees with what I arrived at. Text verbatim (ex. 15) is following
Exercise 15/p36
One urn contains only white balls, while another urn contains 30 white
and 10 black balls. An urn is selected at random and then a ball is
drawn (at random) from the urn. The ball turns out to be white and is
then put back into the urn. What is the probability that another ball
drawn from the same urn will be black?
The answer given is $P = frac{3}{28}$
If the above is correct, my solution must be flawed somehow:
Let $P(x_k)$ be the probability that ball $x$ is drawn in $k-th$ draw.
Then we seek the probability $P(b_2|w_1)$. Given that urns were selected at random, and we don't which one we ended up with
$$P(b_2|w_1) = P(b_2|w_1|u_1)P(u_1) + P(b_2|w_1|u_2) P(u_2)$$
where $P(u_i) = frac{1}{2}$ is the probability that $i-th$ urn was selected. The first of these terms vanishes, as urn 1 does not contain any black balls, hence the probability of seeing a black ball is 0.
For the second, as the white ball is put back into the urn, the second draw is obviously independent with the first, hence
$$P(b_2|w_1|u_2) = frac{1}{4}$$
Thus the answer would come to be
$$P(b_2|w_1) = frac{1}{8}$$
which is obviously very different from given answer.
Where is the mistake?
probability probability-theory
probability probability-theory
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
edited Dec 11 '18 at 4:59
Mike Pierce
11.5k103584
11.5k103584
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
asked Dec 11 '18 at 1:31
Teremin12Teremin12
32
32
add a comment |
add a comment |
1 Answer
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Let $U_1$ is the event ball drawn from Urn-$1$.
Let $U_2$ is the event ball drawn from Urn-$2$.
Let $W$ be the event that the ball is white.
$$P(U_1|W)=dfrac{P(W|U_1)cdot P(U_1)}{P(W|U_1)cdot P(U_1)+P(W|U_2)cdot P(U_2)}=dfrac{1cdotdfrac12}{1cdotdfrac12+dfrac34cdotdfrac12}=dfrac47$$
From this we can say the $P(U_2|W)=dfrac37$
The probability that the white ball is drawn second time is $$dfrac47cdot1+dfrac37cdotdfrac34=dfrac{25}{28}$$
Therefore, the probability that the black ball is drawn second time $=1-dfrac{25}{28}=dfrac{3}{28}$
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $U_1$ is the event ball drawn from Urn-$1$.
Let $U_2$ is the event ball drawn from Urn-$2$.
Let $W$ be the event that the ball is white.
$$P(U_1|W)=dfrac{P(W|U_1)cdot P(U_1)}{P(W|U_1)cdot P(U_1)+P(W|U_2)cdot P(U_2)}=dfrac{1cdotdfrac12}{1cdotdfrac12+dfrac34cdotdfrac12}=dfrac47$$
From this we can say the $P(U_2|W)=dfrac37$
The probability that the white ball is drawn second time is $$dfrac47cdot1+dfrac37cdotdfrac34=dfrac{25}{28}$$
Therefore, the probability that the black ball is drawn second time $=1-dfrac{25}{28}=dfrac{3}{28}$
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
add a comment |
$begingroup$
Let $U_1$ is the event ball drawn from Urn-$1$.
Let $U_2$ is the event ball drawn from Urn-$2$.
Let $W$ be the event that the ball is white.
$$P(U_1|W)=dfrac{P(W|U_1)cdot P(U_1)}{P(W|U_1)cdot P(U_1)+P(W|U_2)cdot P(U_2)}=dfrac{1cdotdfrac12}{1cdotdfrac12+dfrac34cdotdfrac12}=dfrac47$$
From this we can say the $P(U_2|W)=dfrac37$
The probability that the white ball is drawn second time is $$dfrac47cdot1+dfrac37cdotdfrac34=dfrac{25}{28}$$
Therefore, the probability that the black ball is drawn second time $=1-dfrac{25}{28}=dfrac{3}{28}$
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
add a comment |
$begingroup$
Let $U_1$ is the event ball drawn from Urn-$1$.
Let $U_2$ is the event ball drawn from Urn-$2$.
Let $W$ be the event that the ball is white.
$$P(U_1|W)=dfrac{P(W|U_1)cdot P(U_1)}{P(W|U_1)cdot P(U_1)+P(W|U_2)cdot P(U_2)}=dfrac{1cdotdfrac12}{1cdotdfrac12+dfrac34cdotdfrac12}=dfrac47$$
From this we can say the $P(U_2|W)=dfrac37$
The probability that the white ball is drawn second time is $$dfrac47cdot1+dfrac37cdotdfrac34=dfrac{25}{28}$$
Therefore, the probability that the black ball is drawn second time $=1-dfrac{25}{28}=dfrac{3}{28}$
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
$endgroup$
Let $U_1$ is the event ball drawn from Urn-$1$.
Let $U_2$ is the event ball drawn from Urn-$2$.
Let $W$ be the event that the ball is white.
$$P(U_1|W)=dfrac{P(W|U_1)cdot P(U_1)}{P(W|U_1)cdot P(U_1)+P(W|U_2)cdot P(U_2)}=dfrac{1cdotdfrac12}{1cdotdfrac12+dfrac34cdotdfrac12}=dfrac47$$
From this we can say the $P(U_2|W)=dfrac37$
The probability that the white ball is drawn second time is $$dfrac47cdot1+dfrac37cdotdfrac34=dfrac{25}{28}$$
Therefore, the probability that the black ball is drawn second time $=1-dfrac{25}{28}=dfrac{3}{28}$
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
answered Dec 11 '18 at 1:40
Key FlexKey Flex
8,28261233
8,28261233
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
add a comment |
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
Alternatively, the probability that the black ball is drawn second time is $dfrac47cdot0+dfrac37cdotdfrac14=dfrac{3}{28}$
$endgroup$
– Henry
Dec 11 '18 at 1:42
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it?
$endgroup$
– Teremin12
Dec 11 '18 at 1:57
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
$begingroup$
@Teremin12 Yes.
$endgroup$
– Key Flex
Dec 11 '18 at 2:00
add a comment |
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