If we extract $3$ marbles randomly from an urn that contains…drawing balls from an urn (conditional probability)Extracting a ball from an urn, introducing it into the second. Expected value=?Probability, picking balls from urnAn urn contains 3 white balls and 4 black balls. Second urn contains 6 white balls and 4 black balls.Urn I contains 6 whites and 4 blacks balls. Urn II contains 2 white and 2 black balls.Probability: A flaw in logic? The emperor's proposition with marbles and two urnsExtraction of marbles from a boxDrawing 4 balls from an urn without replacement and a bonus ballExtract balls from urn probabilityIf I can solve this simple urn probability problem, what do I need to solve this more complex urn problem?
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If we extract $3$ marbles randomly from an urn that contains…
drawing balls from an urn (conditional probability)Extracting a ball from an urn, introducing it into the second. Expected value=?Probability, picking balls from urnAn urn contains 3 white balls and 4 black balls. Second urn contains 6 white balls and 4 black balls.Urn I contains 6 whites and 4 blacks balls. Urn II contains 2 white and 2 black balls.Probability: A flaw in logic? The emperor's proposition with marbles and two urnsExtraction of marbles from a boxDrawing 4 balls from an urn without replacement and a bonus ballExtract balls from urn probabilityIf I can solve this simple urn probability problem, what do I need to solve this more complex urn problem?
$begingroup$
If we extract $3$ marbles randomly from an urn that contains $6$ white marbles and $5$ black marbles, what is the probability that one is white and the other two black?
Solution. If we take into account the order of extraction and suppose that every possible outcome of the sample space is equiprobable, the probability required is
$frac(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)11 · 10 · 9=frac3 · 120990=4/
11= 0.3636$
I don't understand the numerator $(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)$. The denominator $11 · 10 · 9$ are the possible cases, you have $11$ possibilities for the first marbles, $10$ for the second, $9$ for the last one.
probability statistics
asked Mar 17 at 11:55
user649882user649882
133
$endgroup$
add a comment |
$begingroup$
If we extract $3$ marbles randomly from an urn that contains $6$ white marbles and $5$ black marbles, what is the probability that one is white and the other two black?
Solution. If we take into account the order of extraction and suppose that every possible outcome of the sample space is equiprobable, the probability required is
$frac(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)11 · 10 · 9=frac3 · 120990=4/
11= 0.3636$
I don't understand the numerator $(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)$. The denominator $11 · 10 · 9$ are the possible cases, you have $11$ possibilities for the first marbles, $10$ for the second, $9$ for the last one.
probability statistics
asked Mar 17 at 11:55
user649882user649882
133
$endgroup$
add a comment |
$begingroup$
If we extract $3$ marbles randomly from an urn that contains $6$ white marbles and $5$ black marbles, what is the probability that one is white and the other two black?
Solution. If we take into account the order of extraction and suppose that every possible outcome of the sample space is equiprobable, the probability required is
$frac(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)11 · 10 · 9=frac3 · 120990=4/
11= 0.3636$
I don't understand the numerator $(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)$. The denominator $11 · 10 · 9$ are the possible cases, you have $11$ possibilities for the first marbles, $10$ for the second, $9$ for the last one.
probability statistics
asked Mar 17 at 11:55
user649882user649882
133
$endgroup$
If we extract $3$ marbles randomly from an urn that contains $6$ white marbles and $5$ black marbles, what is the probability that one is white and the other two black?
Solution. If we take into account the order of extraction and suppose that every possible outcome of the sample space is equiprobable, the probability required is
$frac(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)11 · 10 · 9=frac3 · 120990=4/
11= 0.3636$
I don't understand the numerator $(6 · 5 · 4) + (5 · 6 · 4) + (5 · 4 · 6)$. The denominator $11 · 10 · 9$ are the possible cases, you have $11$ possibilities for the first marbles, $10$ for the second, $9$ for the last one.
probability statistics
probability statistics
asked Mar 17 at 11:55
user649882user649882
133
asked Mar 17 at 11:55
user649882user649882
133
asked Mar 17 at 11:55
user649882user649882
133
asked Mar 17 at 11:55
user649882user649882
133
asked Mar 17 at 11:55
user649882user649882
133
133
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.
This with:
- $P(WBB)=frac611frac510frac49$
- $P(BWB)=frac511frac610frac49$
- $P(BBW)=frac511frac410frac69$
More precise we can write: $$P(BWB)=P(B_1cap W_2cap B_3)=P(B_1)P(W_2mid B_1)P(B_3mid B_1cap W_2)=frac511frac610frac49$$where e.g. $W_2$ stands for the event that the second chosen marble is white.
Addition of these probabilities gives you the probability of the mentioned event.
answered Mar 17 at 12:03
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
The numerator is the number of possible ways to end up with $1$ white and $2$ blacks balls.
To see that, notice that you can draw the one white ball either as the first, or as the second, or as the third ball.
If you draw it as the first ball, then since there are $6$ white balls, you have $6$ ways of drawing a white ball first. After that, you need to draw two black balls, and there are 5 possible black balls for the first black ball, and after that 4 black balls for the second black ball. So: $6 times 5 times 4$ possibilities of drawing one white ball and two black balls while drawing the white balls first.
Likewise, the $5 times 6 times 4$ is the number of ways of drawing one white ball and two black balls while drawing the one white ball as the second ball. And the $5 times 4 times 6$ is the number of ways to draw one white ball and two black balls while drawing the one white ball as the third ball.
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6times 5times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.
This with:
- $P(WBB)=frac611frac510frac49$
- $P(BWB)=frac511frac610frac49$
- $P(BBW)=frac511frac410frac69$
More precise we can write: $$P(BWB)=P(B_1cap W_2cap B_3)=P(B_1)P(W_2mid B_1)P(B_3mid B_1cap W_2)=frac511frac610frac49$$where e.g. $W_2$ stands for the event that the second chosen marble is white.
Addition of these probabilities gives you the probability of the mentioned event.
answered Mar 17 at 12:03
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.
This with:
- $P(WBB)=frac611frac510frac49$
- $P(BWB)=frac511frac610frac49$
- $P(BBW)=frac511frac410frac69$
More precise we can write: $$P(BWB)=P(B_1cap W_2cap B_3)=P(B_1)P(W_2mid B_1)P(B_3mid B_1cap W_2)=frac511frac610frac49$$where e.g. $W_2$ stands for the event that the second chosen marble is white.
Addition of these probabilities gives you the probability of the mentioned event.
answered Mar 17 at 12:03
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.
This with:
- $P(WBB)=frac611frac510frac49$
- $P(BWB)=frac511frac610frac49$
- $P(BBW)=frac511frac410frac69$
More precise we can write: $$P(BWB)=P(B_1cap W_2cap B_3)=P(B_1)P(W_2mid B_1)P(B_3mid B_1cap W_2)=frac511frac610frac49$$where e.g. $W_2$ stands for the event that the second chosen marble is white.
Addition of these probabilities gives you the probability of the mentioned event.
answered Mar 17 at 12:03
drhabdrhab
103k545136
$endgroup$
There are $3$ mutual exclusive possibilities: WBB, BWB, BBW.
This with:
- $P(WBB)=frac611frac510frac49$
- $P(BWB)=frac511frac610frac49$
- $P(BBW)=frac511frac410frac69$
More precise we can write: $$P(BWB)=P(B_1cap W_2cap B_3)=P(B_1)P(W_2mid B_1)P(B_3mid B_1cap W_2)=frac511frac610frac49$$where e.g. $W_2$ stands for the event that the second chosen marble is white.
Addition of these probabilities gives you the probability of the mentioned event.
answered Mar 17 at 12:03
drhabdrhab
103k545136
edited Mar 17 at 12:08
answered Mar 17 at 12:03
drhabdrhab
103k545136
answered Mar 17 at 12:03
drhabdrhab
103k545136
answered Mar 17 at 12:03
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
The numerator is the number of possible ways to end up with $1$ white and $2$ blacks balls.
To see that, notice that you can draw the one white ball either as the first, or as the second, or as the third ball.
If you draw it as the first ball, then since there are $6$ white balls, you have $6$ ways of drawing a white ball first. After that, you need to draw two black balls, and there are 5 possible black balls for the first black ball, and after that 4 black balls for the second black ball. So: $6 times 5 times 4$ possibilities of drawing one white ball and two black balls while drawing the white balls first.
Likewise, the $5 times 6 times 4$ is the number of ways of drawing one white ball and two black balls while drawing the one white ball as the second ball. And the $5 times 4 times 6$ is the number of ways to draw one white ball and two black balls while drawing the one white ball as the third ball.
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
The numerator is the number of possible ways to end up with $1$ white and $2$ blacks balls.
To see that, notice that you can draw the one white ball either as the first, or as the second, or as the third ball.
If you draw it as the first ball, then since there are $6$ white balls, you have $6$ ways of drawing a white ball first. After that, you need to draw two black balls, and there are 5 possible black balls for the first black ball, and after that 4 black balls for the second black ball. So: $6 times 5 times 4$ possibilities of drawing one white ball and two black balls while drawing the white balls first.
Likewise, the $5 times 6 times 4$ is the number of ways of drawing one white ball and two black balls while drawing the one white ball as the second ball. And the $5 times 4 times 6$ is the number of ways to draw one white ball and two black balls while drawing the one white ball as the third ball.
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
The numerator is the number of possible ways to end up with $1$ white and $2$ blacks balls.
To see that, notice that you can draw the one white ball either as the first, or as the second, or as the third ball.
If you draw it as the first ball, then since there are $6$ white balls, you have $6$ ways of drawing a white ball first. After that, you need to draw two black balls, and there are 5 possible black balls for the first black ball, and after that 4 black balls for the second black ball. So: $6 times 5 times 4$ possibilities of drawing one white ball and two black balls while drawing the white balls first.
Likewise, the $5 times 6 times 4$ is the number of ways of drawing one white ball and two black balls while drawing the one white ball as the second ball. And the $5 times 4 times 6$ is the number of ways to draw one white ball and two black balls while drawing the one white ball as the third ball.
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
$endgroup$
The numerator is the number of possible ways to end up with $1$ white and $2$ blacks balls.
To see that, notice that you can draw the one white ball either as the first, or as the second, or as the third ball.
If you draw it as the first ball, then since there are $6$ white balls, you have $6$ ways of drawing a white ball first. After that, you need to draw two black balls, and there are 5 possible black balls for the first black ball, and after that 4 black balls for the second black ball. So: $6 times 5 times 4$ possibilities of drawing one white ball and two black balls while drawing the white balls first.
Likewise, the $5 times 6 times 4$ is the number of ways of drawing one white ball and two black balls while drawing the one white ball as the second ball. And the $5 times 4 times 6$ is the number of ways to draw one white ball and two black balls while drawing the one white ball as the third ball.
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
edited Mar 17 at 13:43
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
answered Mar 17 at 12:03
Bram28Bram28
63.9k44793
63.9k44793
add a comment |
add a comment |
$begingroup$
They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6times 5times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6times 5times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6times 5times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
They are considering the three possible orders of selecting one white and two blacks: $WBB, BWB$ and $BBW$. E.g. for $WBB$ (first marble you pick is white, then next two are black), the number of ways to do this is $6times 5times 4$, since you can pick the white marble in $6$ ways, then pick a black marble in $5$ ways, then pick the second black marble in $4$ ways.
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 17 at 11:59
Minus One-TwelfthMinus One-Twelfth
2,823413
2,823413
add a comment |
add a comment |
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