Mixing
Probability Distributions (Tree Diagram)
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Satish picks a card at random from an ordinary pack. If the card is ace, he stops; if not, he continues to pick cards at random, without replacement, until either an ace is picked, or four cards have been drawn. The random variable C is the total number of cards drawn. Construct a tree diagram to illustrate the possible outcomes of the experiment, and use it to calculate the probability distribution of C.
Can anyone explain this question for me? Thanks.
probability probability-distributions
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
$endgroup$
add a comment |
$begingroup$
Satish picks a card at random from an ordinary pack. If the card is ace, he stops; if not, he continues to pick cards at random, without replacement, until either an ace is picked, or four cards have been drawn. The random variable C is the total number of cards drawn. Construct a tree diagram to illustrate the possible outcomes of the experiment, and use it to calculate the probability distribution of C.
Can anyone explain this question for me? Thanks.
probability probability-distributions
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
$endgroup$
$begingroup$
This is not (probability-distributions).
$endgroup$
– Brian Tung
Aug 25 '15 at 16:38
add a comment |
$begingroup$
Satish picks a card at random from an ordinary pack. If the card is ace, he stops; if not, he continues to pick cards at random, without replacement, until either an ace is picked, or four cards have been drawn. The random variable C is the total number of cards drawn. Construct a tree diagram to illustrate the possible outcomes of the experiment, and use it to calculate the probability distribution of C.
Can anyone explain this question for me? Thanks.
probability probability-distributions
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
$endgroup$
Satish picks a card at random from an ordinary pack. If the card is ace, he stops; if not, he continues to pick cards at random, without replacement, until either an ace is picked, or four cards have been drawn. The random variable C is the total number of cards drawn. Construct a tree diagram to illustrate the possible outcomes of the experiment, and use it to calculate the probability distribution of C.
Can anyone explain this question for me? Thanks.
probability probability-distributions
probability probability-distributions
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
asked Aug 25 '15 at 16:33
MathxxMathxx
3,41811544
3,41811544
$begingroup$
This is not (probability-distributions).
$endgroup$
– Brian Tung
Aug 25 '15 at 16:38
add a comment |
$begingroup$
This is not (probability-distributions).
$endgroup$
– Brian Tung
Aug 25 '15 at 16:38
$begingroup$
This is not (probability-distributions).
$endgroup$
– Brian Tung
Aug 25 '15 at 16:38
$begingroup$
This is not (probability-distributions).
$endgroup$
– Brian Tung
Aug 25 '15 at 16:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Perhaps it would be easier to write it out as a piece of code—viz.,
if first card is an ace
C = 1
else if second card is an ace
C = 2
else if third card is an ace
C = 3
else
C = 4
Now, draw the above as a tree. Start with a "$52$ cards, $4$ aces" node, and have two branches coming out of it: one for ace, and one for not-ace. The ace branch leads to a node with $C = 1$; this node is a leaf. The not-ace branch leads to another node "$51$ cards, $4$ aces" with two branches: one for ace, and one for not-ace. The ace branch leads to a node with $C = 2$; this node is a leaf. Again, the not-ace branch leads to another node with two branches. And so on.
Nexct, label each branch with its associated probability. For instance, the first two branches have probability $4/52 = 1/13$ and $48/52 = 12/13$, respectively. Finally, use independence and multiplication to determine the probability of ending up at each leaf node.
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
$endgroup$
add a comment |
$begingroup$
You are being asked to construct a Decision tree based on wether you draw an ace or not.
You will have four nodes, for one, two, three, or four cards drawn, as well as 'start' and 'stop'. The decision after each node is "Was the last card an ace?"; if "yes", report the count then go to 'stop', if "no" then move to the next draw. But automatically report "4" and go to 'stop' if you reach the fourth draw.
To calculate the probability of each result, evaluate the probability of each decision's answer needed to reach that result, and multiply.
For bonus points, instead of a chain, include a loop decision in the process.
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Perhaps it would be easier to write it out as a piece of code—viz.,
if first card is an ace
C = 1
else if second card is an ace
C = 2
else if third card is an ace
C = 3
else
C = 4
Now, draw the above as a tree. Start with a "$52$ cards, $4$ aces" node, and have two branches coming out of it: one for ace, and one for not-ace. The ace branch leads to a node with $C = 1$; this node is a leaf. The not-ace branch leads to another node "$51$ cards, $4$ aces" with two branches: one for ace, and one for not-ace. The ace branch leads to a node with $C = 2$; this node is a leaf. Again, the not-ace branch leads to another node with two branches. And so on.
Nexct, label each branch with its associated probability. For instance, the first two branches have probability $4/52 = 1/13$ and $48/52 = 12/13$, respectively. Finally, use independence and multiplication to determine the probability of ending up at each leaf node.
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
$endgroup$
add a comment |
$begingroup$
Perhaps it would be easier to write it out as a piece of code—viz.,
if first card is an ace
C = 1
else if second card is an ace
C = 2
else if third card is an ace
C = 3
else
C = 4
Now, draw the above as a tree. Start with a "$52$ cards, $4$ aces" node, and have two branches coming out of it: one for ace, and one for not-ace. The ace branch leads to a node with $C = 1$; this node is a leaf. The not-ace branch leads to another node "$51$ cards, $4$ aces" with two branches: one for ace, and one for not-ace. The ace branch leads to a node with $C = 2$; this node is a leaf. Again, the not-ace branch leads to another node with two branches. And so on.
Nexct, label each branch with its associated probability. For instance, the first two branches have probability $4/52 = 1/13$ and $48/52 = 12/13$, respectively. Finally, use independence and multiplication to determine the probability of ending up at each leaf node.
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
$endgroup$
add a comment |
$begingroup$
Perhaps it would be easier to write it out as a piece of code—viz.,
if first card is an ace
C = 1
else if second card is an ace
C = 2
else if third card is an ace
C = 3
else
C = 4
Now, draw the above as a tree. Start with a "$52$ cards, $4$ aces" node, and have two branches coming out of it: one for ace, and one for not-ace. The ace branch leads to a node with $C = 1$; this node is a leaf. The not-ace branch leads to another node "$51$ cards, $4$ aces" with two branches: one for ace, and one for not-ace. The ace branch leads to a node with $C = 2$; this node is a leaf. Again, the not-ace branch leads to another node with two branches. And so on.
Nexct, label each branch with its associated probability. For instance, the first two branches have probability $4/52 = 1/13$ and $48/52 = 12/13$, respectively. Finally, use independence and multiplication to determine the probability of ending up at each leaf node.
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
$endgroup$
Perhaps it would be easier to write it out as a piece of code—viz.,
if first card is an ace
C = 1
else if second card is an ace
C = 2
else if third card is an ace
C = 3
else
C = 4
Now, draw the above as a tree. Start with a "$52$ cards, $4$ aces" node, and have two branches coming out of it: one for ace, and one for not-ace. The ace branch leads to a node with $C = 1$; this node is a leaf. The not-ace branch leads to another node "$51$ cards, $4$ aces" with two branches: one for ace, and one for not-ace. The ace branch leads to a node with $C = 2$; this node is a leaf. Again, the not-ace branch leads to another node with two branches. And so on.
Nexct, label each branch with its associated probability. For instance, the first two branches have probability $4/52 = 1/13$ and $48/52 = 12/13$, respectively. Finally, use independence and multiplication to determine the probability of ending up at each leaf node.
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
answered Aug 25 '15 at 16:44
Brian TungBrian Tung
26.2k32656
26.2k32656
add a comment |
add a comment |
$begingroup$
You are being asked to construct a Decision tree based on wether you draw an ace or not.
You will have four nodes, for one, two, three, or four cards drawn, as well as 'start' and 'stop'. The decision after each node is "Was the last card an ace?"; if "yes", report the count then go to 'stop', if "no" then move to the next draw. But automatically report "4" and go to 'stop' if you reach the fourth draw.
To calculate the probability of each result, evaluate the probability of each decision's answer needed to reach that result, and multiply.
For bonus points, instead of a chain, include a loop decision in the process.
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
$endgroup$
add a comment |
$begingroup$
You are being asked to construct a Decision tree based on wether you draw an ace or not.
You will have four nodes, for one, two, three, or four cards drawn, as well as 'start' and 'stop'. The decision after each node is "Was the last card an ace?"; if "yes", report the count then go to 'stop', if "no" then move to the next draw. But automatically report "4" and go to 'stop' if you reach the fourth draw.
To calculate the probability of each result, evaluate the probability of each decision's answer needed to reach that result, and multiply.
For bonus points, instead of a chain, include a loop decision in the process.
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
$endgroup$
add a comment |
$begingroup$
You are being asked to construct a Decision tree based on wether you draw an ace or not.
You will have four nodes, for one, two, three, or four cards drawn, as well as 'start' and 'stop'. The decision after each node is "Was the last card an ace?"; if "yes", report the count then go to 'stop', if "no" then move to the next draw. But automatically report "4" and go to 'stop' if you reach the fourth draw.
To calculate the probability of each result, evaluate the probability of each decision's answer needed to reach that result, and multiply.
For bonus points, instead of a chain, include a loop decision in the process.
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
$endgroup$
You are being asked to construct a Decision tree based on wether you draw an ace or not.
You will have four nodes, for one, two, three, or four cards drawn, as well as 'start' and 'stop'. The decision after each node is "Was the last card an ace?"; if "yes", report the count then go to 'stop', if "no" then move to the next draw. But automatically report "4" and go to 'stop' if you reach the fourth draw.
To calculate the probability of each result, evaluate the probability of each decision's answer needed to reach that result, and multiply.
For bonus points, instead of a chain, include a loop decision in the process.
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
answered Aug 26 '15 at 2:08
Graham KempGraham Kemp
87.9k43578
87.9k43578
add a comment |
add a comment |
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$begingroup$
This is not (probability-distributions).
$endgroup$
– Brian Tung
Aug 25 '15 at 16:38