Mixing
Distribution of the heads-tails difference after three coin tosses
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Three fair coins are tossed and D is the positive difference between the number of heads and the number of tails obtained, so D takes the values 1 and 3. Tabulate the probability distribution of D and calculate E(D).
I have made groups like
with difference 1:
(HTT) (TTH) (HTH) (THT) (HHT) (THH)
with difference 3:
(HHH) (TTT)
later I have done
$6 (1/2)^3$ (for the condition where difference is 1)
AND
$2 (1/2)^3$ (for the condition where difference is 3)
The answers I am getting are 3/4 for the first condition and 1/4 for the second condition but I am not sure if this is the right way
probability
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
$endgroup$
add a comment |
$begingroup$
Three fair coins are tossed and D is the positive difference between the number of heads and the number of tails obtained, so D takes the values 1 and 3. Tabulate the probability distribution of D and calculate E(D).
I have made groups like
with difference 1:
(HTT) (TTH) (HTH) (THT) (HHT) (THH)
with difference 3:
(HHH) (TTT)
later I have done
$6 (1/2)^3$ (for the condition where difference is 1)
AND
$2 (1/2)^3$ (for the condition where difference is 3)
The answers I am getting are 3/4 for the first condition and 1/4 for the second condition but I am not sure if this is the right way
probability
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
$endgroup$
3
$begingroup$
What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you.
$endgroup$
– Mark Fischler
Apr 1 '15 at 21:09
1
$begingroup$
I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air.
$endgroup$
– Joffan
Apr 1 '15 at 21:21
1
$begingroup$
Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities.
$endgroup$
– peter.petrov
Apr 1 '15 at 21:31
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oh yes sorry for the mistake
$endgroup$
– Hamza750
Apr 1 '15 at 21:32
$begingroup$
Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D).
$endgroup$
– BruceET
Apr 2 '15 at 0:21
add a comment |
$begingroup$
Three fair coins are tossed and D is the positive difference between the number of heads and the number of tails obtained, so D takes the values 1 and 3. Tabulate the probability distribution of D and calculate E(D).
I have made groups like
with difference 1:
(HTT) (TTH) (HTH) (THT) (HHT) (THH)
with difference 3:
(HHH) (TTT)
later I have done
$6 (1/2)^3$ (for the condition where difference is 1)
AND
$2 (1/2)^3$ (for the condition where difference is 3)
The answers I am getting are 3/4 for the first condition and 1/4 for the second condition but I am not sure if this is the right way
probability
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
$endgroup$
Three fair coins are tossed and D is the positive difference between the number of heads and the number of tails obtained, so D takes the values 1 and 3. Tabulate the probability distribution of D and calculate E(D).
I have made groups like
with difference 1:
(HTT) (TTH) (HTH) (THT) (HHT) (THH)
with difference 3:
(HHH) (TTT)
later I have done
$6 (1/2)^3$ (for the condition where difference is 1)
AND
$2 (1/2)^3$ (for the condition where difference is 3)
The answers I am getting are 3/4 for the first condition and 1/4 for the second condition but I am not sure if this is the right way
probability
probability
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
edited Apr 2 '15 at 1:40
user147263
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
asked Apr 1 '15 at 21:07
Hamza750Hamza750
315
315
3
$begingroup$
What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you.
$endgroup$
– Mark Fischler
Apr 1 '15 at 21:09
1
$begingroup$
I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air.
$endgroup$
– Joffan
Apr 1 '15 at 21:21
1
$begingroup$
Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities.
$endgroup$
– peter.petrov
Apr 1 '15 at 21:31
$begingroup$
oh yes sorry for the mistake
$endgroup$
– Hamza750
Apr 1 '15 at 21:32
$begingroup$
Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D).
$endgroup$
– BruceET
Apr 2 '15 at 0:21
add a comment |
3
$begingroup$
What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you.
$endgroup$
– Mark Fischler
Apr 1 '15 at 21:09
1
$begingroup$
I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air.
$endgroup$
– Joffan
Apr 1 '15 at 21:21
1
$begingroup$
Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities.
$endgroup$
– peter.petrov
Apr 1 '15 at 21:31
$begingroup$
oh yes sorry for the mistake
$endgroup$
– Hamza750
Apr 1 '15 at 21:32
$begingroup$
Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D).
$endgroup$
– BruceET
Apr 2 '15 at 0:21
3
3
$begingroup$
What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you.
$endgroup$
– Mark Fischler
Apr 1 '15 at 21:09
$begingroup$
What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you.
$endgroup$
– Mark Fischler
Apr 1 '15 at 21:09
1
1
$begingroup$
I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air.
$endgroup$
– Joffan
Apr 1 '15 at 21:21
$begingroup$
I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air.
$endgroup$
– Joffan
Apr 1 '15 at 21:21
1
1
$begingroup$
Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities.
$endgroup$
– peter.petrov
Apr 1 '15 at 21:31
$begingroup$
Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities.
$endgroup$
– peter.petrov
Apr 1 '15 at 21:31
$begingroup$
oh yes sorry for the mistake
$endgroup$
– Hamza750
Apr 1 '15 at 21:32
$begingroup$
oh yes sorry for the mistake
$endgroup$
– Hamza750
Apr 1 '15 at 21:32
$begingroup$
Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D).
$endgroup$
– BruceET
Apr 2 '15 at 0:21
$begingroup$
Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D).
$endgroup$
– BruceET
Apr 2 '15 at 0:21
add a comment |
1 Answer
1
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$begingroup$
Your calculation of the probability distribution of the absolute difference is correct:
- $Pr(D=1) = frac34$
- $Pr(D=3) = frac14$
You could have used these to find the expected absolute difference:
- $E[D]=1times frac34 +3 times frac14 = frac32 .$
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
Your calculation of the probability distribution of the absolute difference is correct:
- $Pr(D=1) = frac34$
- $Pr(D=3) = frac14$
You could have used these to find the expected absolute difference:
- $E[D]=1times frac34 +3 times frac14 = frac32 .$
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Your calculation of the probability distribution of the absolute difference is correct:
- $Pr(D=1) = frac34$
- $Pr(D=3) = frac14$
You could have used these to find the expected absolute difference:
- $E[D]=1times frac34 +3 times frac14 = frac32 .$
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
Your calculation of the probability distribution of the absolute difference is correct:
- $Pr(D=1) = frac34$
- $Pr(D=3) = frac14$
You could have used these to find the expected absolute difference:
- $E[D]=1times frac34 +3 times frac14 = frac32 .$
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
$endgroup$
Your calculation of the probability distribution of the absolute difference is correct:
- $Pr(D=1) = frac34$
- $Pr(D=3) = frac14$
You could have used these to find the expected absolute difference:
- $E[D]=1times frac34 +3 times frac14 = frac32 .$
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
answered Jul 27 '15 at 8:46
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
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3
$begingroup$
What have you tried, and where are you having problems? This is not a good site to just have people do your homework for you.
$endgroup$
– Mark Fischler
Apr 1 '15 at 21:09
1
$begingroup$
I would never trust coins I got from a fair. They're bound to be weighted, or have two heads, or split into two mid-air.
$endgroup$
– Joffan
Apr 1 '15 at 21:21
1
$begingroup$
Your answers seem correct. Just the difference there - it's not 0, it's 3. And also you stopped before calculating the expectation, it seems. You only calculated the probabilities.
$endgroup$
– peter.petrov
Apr 1 '15 at 21:31
$begingroup$
oh yes sorry for the mistake
$endgroup$
– Hamza750
Apr 1 '15 at 21:32
$begingroup$
Because there are only 8 outcomes in terms of H's and T's when three coins are tossed, I suppose you were intended to get the distribution of D by direct enumeration--the way you began. Make a distribution table, telling the probability of each value of D. Then use it to find E(D).
$endgroup$
– BruceET
Apr 2 '15 at 0:21