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generate event with $6.75space p^2q$, $20space p^3q^2$, $3.9space pq$?
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Can we generate event with probabilities (a) $6.75space p^2q$, (b) $20space p^3q^2$, (c) $3.9space pq$? If yes, then how? If no, then why?
Here, in a coin toss, let P(occurring head)$=p=1-q$
I can easily generate events for $3space pq$(with three trials which is just $3space p^2q+3space pq^2$), $10space p^3q^2$ (with five trials) and $3space p^2q$ (with three trials), but could not find anything in these cases. Thanks for any help!
probability probability-theory
asked Jul 15 at 18:24
Stat_prob_001
18710
add a comment |Â
up vote
2
down vote
favorite
Can we generate event with probabilities (a) $6.75space p^2q$, (b) $20space p^3q^2$, (c) $3.9space pq$? If yes, then how? If no, then why?
Here, in a coin toss, let P(occurring head)$=p=1-q$
I can easily generate events for $3space pq$(with three trials which is just $3space p^2q+3space pq^2$), $10space p^3q^2$ (with five trials) and $3space p^2q$ (with three trials), but could not find anything in these cases. Thanks for any help!
probability probability-theory
asked Jul 15 at 18:24
Stat_prob_001
18710
1
See also this more general question that was inspired by this question.
– joriki
Jul 16 at 21:40
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Can we generate event with probabilities (a) $6.75space p^2q$, (b) $20space p^3q^2$, (c) $3.9space pq$? If yes, then how? If no, then why?
Here, in a coin toss, let P(occurring head)$=p=1-q$
I can easily generate events for $3space pq$(with three trials which is just $3space p^2q+3space pq^2$), $10space p^3q^2$ (with five trials) and $3space p^2q$ (with three trials), but could not find anything in these cases. Thanks for any help!
probability probability-theory
asked Jul 15 at 18:24
Stat_prob_001
18710
Can we generate event with probabilities (a) $6.75space p^2q$, (b) $20space p^3q^2$, (c) $3.9space pq$? If yes, then how? If no, then why?
Here, in a coin toss, let P(occurring head)$=p=1-q$
I can easily generate events for $3space pq$(with three trials which is just $3space p^2q+3space pq^2$), $10space p^3q^2$ (with five trials) and $3space p^2q$ (with three trials), but could not find anything in these cases. Thanks for any help!
probability probability-theory
asked Jul 15 at 18:24
Stat_prob_001
18710
asked Jul 15 at 18:24
Stat_prob_001
18710
asked Jul 15 at 18:24
Stat_prob_001
18710
asked Jul 15 at 18:24
Stat_prob_001
18710
18710
1
See also this more general question that was inspired by this question.
– joriki
Jul 16 at 21:40
add a comment |Â
1
See also this more general question that was inspired by this question.
– joriki
Jul 16 at 21:40
1
1
See also this more general question that was inspired by this question.
– joriki
Jul 16 at 21:40
See also this more general question that was inspired by this question.
– joriki
Jul 16 at 21:40
add a comment |Â
1 Answer
1
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oldest
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up vote
1
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For (b), for $0le kle4$ choose $20binom4k$ patterns with $3+k$ heads and $2+(4-k)$ tails, and thus each with probability $p^3+kq^2+(4-k)$. This is possible because $binom93+kge20binom4k$ for all $k$. The sum of the probabilities is
$$
sum_k=0^420binom4kp^3+kq^2+(4-k)=20p^3q^2(p+q)^4=20p^3q^2;.
$$
You can't do the other two for general $p$. Consider $p$ the reciprocal of any odd prime. Then the denominator of any probability in any finite number of trials is always a power of that odd prime and can never contain a factor of $2$, whereas the denominators of the two probabilities in $(a)$ and $(c)$ do.
This of course assumes that you were implying that the problem should be solved with a finite number of trials. It's easy to achieve any desired probability with infinitely many trials.
answered Jul 16 at 4:49
joriki
164k10180328
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For (b), for $0le kle4$ choose $20binom4k$ patterns with $3+k$ heads and $2+(4-k)$ tails, and thus each with probability $p^3+kq^2+(4-k)$. This is possible because $binom93+kge20binom4k$ for all $k$. The sum of the probabilities is
$$
sum_k=0^420binom4kp^3+kq^2+(4-k)=20p^3q^2(p+q)^4=20p^3q^2;.
$$
You can't do the other two for general $p$. Consider $p$ the reciprocal of any odd prime. Then the denominator of any probability in any finite number of trials is always a power of that odd prime and can never contain a factor of $2$, whereas the denominators of the two probabilities in $(a)$ and $(c)$ do.
This of course assumes that you were implying that the problem should be solved with a finite number of trials. It's easy to achieve any desired probability with infinitely many trials.
answered Jul 16 at 4:49
joriki
164k10180328
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
add a comment |Â
up vote
1
down vote
accepted
For (b), for $0le kle4$ choose $20binom4k$ patterns with $3+k$ heads and $2+(4-k)$ tails, and thus each with probability $p^3+kq^2+(4-k)$. This is possible because $binom93+kge20binom4k$ for all $k$. The sum of the probabilities is
$$
sum_k=0^420binom4kp^3+kq^2+(4-k)=20p^3q^2(p+q)^4=20p^3q^2;.
$$
You can't do the other two for general $p$. Consider $p$ the reciprocal of any odd prime. Then the denominator of any probability in any finite number of trials is always a power of that odd prime and can never contain a factor of $2$, whereas the denominators of the two probabilities in $(a)$ and $(c)$ do.
This of course assumes that you were implying that the problem should be solved with a finite number of trials. It's easy to achieve any desired probability with infinitely many trials.
answered Jul 16 at 4:49
joriki
164k10180328
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For (b), for $0le kle4$ choose $20binom4k$ patterns with $3+k$ heads and $2+(4-k)$ tails, and thus each with probability $p^3+kq^2+(4-k)$. This is possible because $binom93+kge20binom4k$ for all $k$. The sum of the probabilities is
$$
sum_k=0^420binom4kp^3+kq^2+(4-k)=20p^3q^2(p+q)^4=20p^3q^2;.
$$
You can't do the other two for general $p$. Consider $p$ the reciprocal of any odd prime. Then the denominator of any probability in any finite number of trials is always a power of that odd prime and can never contain a factor of $2$, whereas the denominators of the two probabilities in $(a)$ and $(c)$ do.
This of course assumes that you were implying that the problem should be solved with a finite number of trials. It's easy to achieve any desired probability with infinitely many trials.
answered Jul 16 at 4:49
joriki
164k10180328
For (b), for $0le kle4$ choose $20binom4k$ patterns with $3+k$ heads and $2+(4-k)$ tails, and thus each with probability $p^3+kq^2+(4-k)$. This is possible because $binom93+kge20binom4k$ for all $k$. The sum of the probabilities is
$$
sum_k=0^420binom4kp^3+kq^2+(4-k)=20p^3q^2(p+q)^4=20p^3q^2;.
$$
You can't do the other two for general $p$. Consider $p$ the reciprocal of any odd prime. Then the denominator of any probability in any finite number of trials is always a power of that odd prime and can never contain a factor of $2$, whereas the denominators of the two probabilities in $(a)$ and $(c)$ do.
This of course assumes that you were implying that the problem should be solved with a finite number of trials. It's easy to achieve any desired probability with infinitely many trials.
answered Jul 16 at 4:49
joriki
164k10180328
edited Jul 16 at 19:41
answered Jul 16 at 4:49
joriki
164k10180328
answered Jul 16 at 4:49
joriki
164k10180328
answered Jul 16 at 4:49
joriki
164k10180328
164k10180328
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
add a comment |Â
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
but the first calculation gives $10p^3q^2$
– Stat_prob_001
Jul 16 at 17:47
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: Sorry! I replaced it by a slightly more involved solution that works (unless I've made another mistake...).
– joriki
Jul 16 at 19:41
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
@Stat_prob_001: I posted and answered a more general form of this question here: math.stackexchange.com/questions/2853884.
– joriki
Jul 16 at 21:39
add a comment |Â
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1
See also this more general question that was inspired by this question.
– joriki
Jul 16 at 21:40