Mixing
calculating mean of coin toss
$begingroup$
i am having a problem solving this one:
we toss a fair coin twice, and then we toss a coin one more for each time we got a H. let X denote the number of total H obtained.
what is the mean?
from what i understand, $E[X] = sum_i P(X=x_i)$, so it should be equal to $0*P{x=0} + ... +4*P{x=4}$. the result should be, according to the written answer, 1.5, but i don't get that. what i wrote is right?
probability probability-distributions
asked Jan 28 at 19:03
q123q123
75
$endgroup$
add a comment |
$begingroup$
i am having a problem solving this one:
we toss a fair coin twice, and then we toss a coin one more for each time we got a H. let X denote the number of total H obtained.
what is the mean?
from what i understand, $E[X] = sum_i P(X=x_i)$, so it should be equal to $0*P{x=0} + ... +4*P{x=4}$. the result should be, according to the written answer, 1.5, but i don't get that. what i wrote is right?
probability probability-distributions
asked Jan 28 at 19:03
q123q123
75
$endgroup$
1
$begingroup$
How could the answer be $4$? The only way to get a $4$ is to throw $HH$ on the first round, and then throw $HH$ again. Anything else gets a lower score.
$endgroup$
– lulu
Jan 28 at 19:35
$begingroup$
The expectation is $$sum_{i=0}^4iP(X=i)$$ What you have written(without multiplying the probability by $i$) sums to $1$. I think the next formula, where you get the answer $4$ must be badly garbled, because it doesn't make sense. Try adding some more details so we can see what you're doing wrong.
$endgroup$
– saulspatz
Jan 28 at 19:45
$begingroup$
lol, i did not do a mistake there. i usually write the probability function as P{}, but for some reason when i use it inside the dollar signs(mathjax), it eliminates the {}, so it look as if i wrote =4 lol
$endgroup$
– q123
Jan 28 at 19:55
$begingroup$
by p(x=4) i meant in the scenario, HH was obtained in additional to 2 additional HH, so there could me, in the maximum case, 4 Heads
$endgroup$
– q123
Jan 28 at 19:58
add a comment |
$begingroup$
i am having a problem solving this one:
we toss a fair coin twice, and then we toss a coin one more for each time we got a H. let X denote the number of total H obtained.
what is the mean?
from what i understand, $E[X] = sum_i P(X=x_i)$, so it should be equal to $0*P{x=0} + ... +4*P{x=4}$. the result should be, according to the written answer, 1.5, but i don't get that. what i wrote is right?
probability probability-distributions
asked Jan 28 at 19:03
q123q123
75
$endgroup$
i am having a problem solving this one:
we toss a fair coin twice, and then we toss a coin one more for each time we got a H. let X denote the number of total H obtained.
what is the mean?
from what i understand, $E[X] = sum_i P(X=x_i)$, so it should be equal to $0*P{x=0} + ... +4*P{x=4}$. the result should be, according to the written answer, 1.5, but i don't get that. what i wrote is right?
probability probability-distributions
probability probability-distributions
asked Jan 28 at 19:03
q123q123
75
asked Jan 28 at 19:03
q123q123
75
asked Jan 28 at 19:03
q123q123
75
asked Jan 28 at 19:03
q123q123
75
asked Jan 28 at 19:03
q123q123
75
75
1
$begingroup$
How could the answer be $4$? The only way to get a $4$ is to throw $HH$ on the first round, and then throw $HH$ again. Anything else gets a lower score.
$endgroup$
– lulu
Jan 28 at 19:35
$begingroup$
The expectation is $$sum_{i=0}^4iP(X=i)$$ What you have written(without multiplying the probability by $i$) sums to $1$. I think the next formula, where you get the answer $4$ must be badly garbled, because it doesn't make sense. Try adding some more details so we can see what you're doing wrong.
$endgroup$
– saulspatz
Jan 28 at 19:45
$begingroup$
lol, i did not do a mistake there. i usually write the probability function as P{}, but for some reason when i use it inside the dollar signs(mathjax), it eliminates the {}, so it look as if i wrote =4 lol
$endgroup$
– q123
Jan 28 at 19:55
$begingroup$
by p(x=4) i meant in the scenario, HH was obtained in additional to 2 additional HH, so there could me, in the maximum case, 4 Heads
$endgroup$
– q123
Jan 28 at 19:58
add a comment |
1
$begingroup$
How could the answer be $4$? The only way to get a $4$ is to throw $HH$ on the first round, and then throw $HH$ again. Anything else gets a lower score.
$endgroup$
– lulu
Jan 28 at 19:35
$begingroup$
The expectation is $$sum_{i=0}^4iP(X=i)$$ What you have written(without multiplying the probability by $i$) sums to $1$. I think the next formula, where you get the answer $4$ must be badly garbled, because it doesn't make sense. Try adding some more details so we can see what you're doing wrong.
$endgroup$
– saulspatz
Jan 28 at 19:45
$begingroup$
lol, i did not do a mistake there. i usually write the probability function as P{}, but for some reason when i use it inside the dollar signs(mathjax), it eliminates the {}, so it look as if i wrote =4 lol
$endgroup$
– q123
Jan 28 at 19:55
$begingroup$
by p(x=4) i meant in the scenario, HH was obtained in additional to 2 additional HH, so there could me, in the maximum case, 4 Heads
$endgroup$
– q123
Jan 28 at 19:58
1
1
$begingroup$
How could the answer be $4$? The only way to get a $4$ is to throw $HH$ on the first round, and then throw $HH$ again. Anything else gets a lower score.
$endgroup$
– lulu
Jan 28 at 19:35
$begingroup$
How could the answer be $4$? The only way to get a $4$ is to throw $HH$ on the first round, and then throw $HH$ again. Anything else gets a lower score.
$endgroup$
– lulu
Jan 28 at 19:35
$begingroup$
The expectation is $$sum_{i=0}^4iP(X=i)$$ What you have written(without multiplying the probability by $i$) sums to $1$. I think the next formula, where you get the answer $4$ must be badly garbled, because it doesn't make sense. Try adding some more details so we can see what you're doing wrong.
$endgroup$
– saulspatz
Jan 28 at 19:45
$begingroup$
The expectation is $$sum_{i=0}^4iP(X=i)$$ What you have written(without multiplying the probability by $i$) sums to $1$. I think the next formula, where you get the answer $4$ must be badly garbled, because it doesn't make sense. Try adding some more details so we can see what you're doing wrong.
$endgroup$
– saulspatz
Jan 28 at 19:45
$begingroup$
lol, i did not do a mistake there. i usually write the probability function as P{}, but for some reason when i use it inside the dollar signs(mathjax), it eliminates the {}, so it look as if i wrote =4 lol
$endgroup$
– q123
Jan 28 at 19:55
$begingroup$
lol, i did not do a mistake there. i usually write the probability function as P{}, but for some reason when i use it inside the dollar signs(mathjax), it eliminates the {}, so it look as if i wrote =4 lol
$endgroup$
– q123
Jan 28 at 19:55
$begingroup$
by p(x=4) i meant in the scenario, HH was obtained in additional to 2 additional HH, so there could me, in the maximum case, 4 Heads
$endgroup$
– q123
Jan 28 at 19:58
$begingroup$
by p(x=4) i meant in the scenario, HH was obtained in additional to 2 additional HH, so there could me, in the maximum case, 4 Heads
$endgroup$
– q123
Jan 28 at 19:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If 0 heads come up (w.p 0.25), $X = 0$ w.p. 0.25 (because no more coins are tossed)
If 1 head comes up (w.p. 0.5), $X=2$ w.p. 0.25 and $X=1$ w.p. 0.25 (because the extra toss is equally likely to come up head or tail)
If 2 heads come up (w.p. 0.25), $X=2$ w.p. 0.0625, $X=3$ w.p. 0.125, and $X=4$ w.p. 0.0625 (the two extra tosses can result in TT, HT, TH, or HH)
Combining it all,
$P(X=0) = 0.25$, $P(X=1) = 0.25$, $P(X=2) = 0.3125$, $P(X=3) = 0.125$, and $P(X=4) = 0.0625$.
Probabilities add up to 1, so that's a good sign!
$mathbb{E}[X] = 1 times 0.25 + 2 times 0.3125 + 3 times 0.125 + 4 times 0.0625 = 1.5$
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
If you toss $TT$, you get 0 heads for $frac{1}{4}$ prob. If you toss TH or HT, you get to toss one more time which may be H or T. The sample space {THT, THH, HTT, HTH} out of $2^3=8$ ways it can play out. Thus it is for a prob of $frac{2}{8}$, you get one head and for another $frac{2}{8}$ prob, you get two heads. The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $frac{1}{16}$ you get two heads (HHTT), and for a prob of $frac{2}{16}$ you get three heads and for a prob of $frac{1}{16}$, you get four heads.
Thus the required probability $=frac{1}{4}times 0+ frac{2}{8}times 1+frac{2}{8}times 2+frac{1}{16}times 2+frac{2}{16}times 3 +frac{1}{16}times 4$
$$ = frac{3}{2}$$
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
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$begingroup$
If 0 heads come up (w.p 0.25), $X = 0$ w.p. 0.25 (because no more coins are tossed)
If 1 head comes up (w.p. 0.5), $X=2$ w.p. 0.25 and $X=1$ w.p. 0.25 (because the extra toss is equally likely to come up head or tail)
If 2 heads come up (w.p. 0.25), $X=2$ w.p. 0.0625, $X=3$ w.p. 0.125, and $X=4$ w.p. 0.0625 (the two extra tosses can result in TT, HT, TH, or HH)
Combining it all,
$P(X=0) = 0.25$, $P(X=1) = 0.25$, $P(X=2) = 0.3125$, $P(X=3) = 0.125$, and $P(X=4) = 0.0625$.
Probabilities add up to 1, so that's a good sign!
$mathbb{E}[X] = 1 times 0.25 + 2 times 0.3125 + 3 times 0.125 + 4 times 0.0625 = 1.5$
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
If 0 heads come up (w.p 0.25), $X = 0$ w.p. 0.25 (because no more coins are tossed)
If 1 head comes up (w.p. 0.5), $X=2$ w.p. 0.25 and $X=1$ w.p. 0.25 (because the extra toss is equally likely to come up head or tail)
If 2 heads come up (w.p. 0.25), $X=2$ w.p. 0.0625, $X=3$ w.p. 0.125, and $X=4$ w.p. 0.0625 (the two extra tosses can result in TT, HT, TH, or HH)
Combining it all,
$P(X=0) = 0.25$, $P(X=1) = 0.25$, $P(X=2) = 0.3125$, $P(X=3) = 0.125$, and $P(X=4) = 0.0625$.
Probabilities add up to 1, so that's a good sign!
$mathbb{E}[X] = 1 times 0.25 + 2 times 0.3125 + 3 times 0.125 + 4 times 0.0625 = 1.5$
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
If 0 heads come up (w.p 0.25), $X = 0$ w.p. 0.25 (because no more coins are tossed)
If 1 head comes up (w.p. 0.5), $X=2$ w.p. 0.25 and $X=1$ w.p. 0.25 (because the extra toss is equally likely to come up head or tail)
If 2 heads come up (w.p. 0.25), $X=2$ w.p. 0.0625, $X=3$ w.p. 0.125, and $X=4$ w.p. 0.0625 (the two extra tosses can result in TT, HT, TH, or HH)
Combining it all,
$P(X=0) = 0.25$, $P(X=1) = 0.25$, $P(X=2) = 0.3125$, $P(X=3) = 0.125$, and $P(X=4) = 0.0625$.
Probabilities add up to 1, so that's a good sign!
$mathbb{E}[X] = 1 times 0.25 + 2 times 0.3125 + 3 times 0.125 + 4 times 0.0625 = 1.5$
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
$endgroup$
If 0 heads come up (w.p 0.25), $X = 0$ w.p. 0.25 (because no more coins are tossed)
If 1 head comes up (w.p. 0.5), $X=2$ w.p. 0.25 and $X=1$ w.p. 0.25 (because the extra toss is equally likely to come up head or tail)
If 2 heads come up (w.p. 0.25), $X=2$ w.p. 0.0625, $X=3$ w.p. 0.125, and $X=4$ w.p. 0.0625 (the two extra tosses can result in TT, HT, TH, or HH)
Combining it all,
$P(X=0) = 0.25$, $P(X=1) = 0.25$, $P(X=2) = 0.3125$, $P(X=3) = 0.125$, and $P(X=4) = 0.0625$.
Probabilities add up to 1, so that's a good sign!
$mathbb{E}[X] = 1 times 0.25 + 2 times 0.3125 + 3 times 0.125 + 4 times 0.0625 = 1.5$
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
answered Jan 29 at 7:35
Aditya DuaAditya Dua
1,15418
1,15418
add a comment |
add a comment |
$begingroup$
If you toss $TT$, you get 0 heads for $frac{1}{4}$ prob. If you toss TH or HT, you get to toss one more time which may be H or T. The sample space {THT, THH, HTT, HTH} out of $2^3=8$ ways it can play out. Thus it is for a prob of $frac{2}{8}$, you get one head and for another $frac{2}{8}$ prob, you get two heads. The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $frac{1}{16}$ you get two heads (HHTT), and for a prob of $frac{2}{16}$ you get three heads and for a prob of $frac{1}{16}$, you get four heads.
Thus the required probability $=frac{1}{4}times 0+ frac{2}{8}times 1+frac{2}{8}times 2+frac{1}{16}times 2+frac{2}{16}times 3 +frac{1}{16}times 4$
$$ = frac{3}{2}$$
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
add a comment |
$begingroup$
If you toss $TT$, you get 0 heads for $frac{1}{4}$ prob. If you toss TH or HT, you get to toss one more time which may be H or T. The sample space {THT, THH, HTT, HTH} out of $2^3=8$ ways it can play out. Thus it is for a prob of $frac{2}{8}$, you get one head and for another $frac{2}{8}$ prob, you get two heads. The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $frac{1}{16}$ you get two heads (HHTT), and for a prob of $frac{2}{16}$ you get three heads and for a prob of $frac{1}{16}$, you get four heads.
Thus the required probability $=frac{1}{4}times 0+ frac{2}{8}times 1+frac{2}{8}times 2+frac{1}{16}times 2+frac{2}{16}times 3 +frac{1}{16}times 4$
$$ = frac{3}{2}$$
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
add a comment |
$begingroup$
If you toss $TT$, you get 0 heads for $frac{1}{4}$ prob. If you toss TH or HT, you get to toss one more time which may be H or T. The sample space {THT, THH, HTT, HTH} out of $2^3=8$ ways it can play out. Thus it is for a prob of $frac{2}{8}$, you get one head and for another $frac{2}{8}$ prob, you get two heads. The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $frac{1}{16}$ you get two heads (HHTT), and for a prob of $frac{2}{16}$ you get three heads and for a prob of $frac{1}{16}$, you get four heads.
Thus the required probability $=frac{1}{4}times 0+ frac{2}{8}times 1+frac{2}{8}times 2+frac{1}{16}times 2+frac{2}{16}times 3 +frac{1}{16}times 4$
$$ = frac{3}{2}$$
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
If you toss $TT$, you get 0 heads for $frac{1}{4}$ prob. If you toss TH or HT, you get to toss one more time which may be H or T. The sample space {THT, THH, HTT, HTH} out of $2^3=8$ ways it can play out. Thus it is for a prob of $frac{2}{8}$, you get one head and for another $frac{2}{8}$ prob, you get two heads. The last case is if you toss HH, you get two more tosses which gives you a favorable outcome of (HHTH, HHHT, HHTT, HHHH} out of $2^4=16$ ways it can play out. Thus it is for a prob of $frac{1}{16}$ you get two heads (HHTT), and for a prob of $frac{2}{16}$ you get three heads and for a prob of $frac{1}{16}$, you get four heads.
Thus the required probability $=frac{1}{4}times 0+ frac{2}{8}times 1+frac{2}{8}times 2+frac{1}{16}times 2+frac{2}{16}times 3 +frac{1}{16}times 4$
$$ = frac{3}{2}$$
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 29 at 8:38
Satish RamanathanSatish Ramanathan
10k31323
10k31323
add a comment |
add a comment |
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1
$begingroup$
How could the answer be $4$? The only way to get a $4$ is to throw $HH$ on the first round, and then throw $HH$ again. Anything else gets a lower score.
$endgroup$
– lulu
Jan 28 at 19:35
$begingroup$
The expectation is $$sum_{i=0}^4iP(X=i)$$ What you have written(without multiplying the probability by $i$) sums to $1$. I think the next formula, where you get the answer $4$ must be badly garbled, because it doesn't make sense. Try adding some more details so we can see what you're doing wrong.
$endgroup$
– saulspatz
Jan 28 at 19:45
$begingroup$
lol, i did not do a mistake there. i usually write the probability function as P{}, but for some reason when i use it inside the dollar signs(mathjax), it eliminates the {}, so it look as if i wrote =4 lol
$endgroup$
– q123
Jan 28 at 19:55
$begingroup$
by p(x=4) i meant in the scenario, HH was obtained in additional to 2 additional HH, so there could me, in the maximum case, 4 Heads
$endgroup$
– q123
Jan 28 at 19:58