Probability that a randomly generated bit string of length 10 begins with a 1 or ends with a 00, given 0 and...
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So, I know that the general formula to find the probability of independent Bernoulli events is:
$C(n,k)p^kq^{n-k}$
But there are a few questions I have about the application of this theorem to this problem. First off, even though we have 10 indices, we are only concerned with 3 of the options. So I don't know how this formula would apply. Also, I am not sure how to account for the or in this scenario. Is it just that the bit beginning with a 1 is just going to be $frac{1}{2}$ and the one for the 00 a the end is just going to be $frac{1}{2} * frac{1}{2}$?
probability probability-theory
edited Nov 15 at 22:05
David G. Stork
9,11821232
asked Nov 15 at 21:40
Jersey Fonseca
356
add a comment |
up vote
0
down vote
favorite
So, I know that the general formula to find the probability of independent Bernoulli events is:
$C(n,k)p^kq^{n-k}$
But there are a few questions I have about the application of this theorem to this problem. First off, even though we have 10 indices, we are only concerned with 3 of the options. So I don't know how this formula would apply. Also, I am not sure how to account for the or in this scenario. Is it just that the bit beginning with a 1 is just going to be $frac{1}{2}$ and the one for the 00 a the end is just going to be $frac{1}{2} * frac{1}{2}$?
probability probability-theory
edited Nov 15 at 22:05
David G. Stork
9,11821232
asked Nov 15 at 21:40
Jersey Fonseca
356
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So, I know that the general formula to find the probability of independent Bernoulli events is:
$C(n,k)p^kq^{n-k}$
But there are a few questions I have about the application of this theorem to this problem. First off, even though we have 10 indices, we are only concerned with 3 of the options. So I don't know how this formula would apply. Also, I am not sure how to account for the or in this scenario. Is it just that the bit beginning with a 1 is just going to be $frac{1}{2}$ and the one for the 00 a the end is just going to be $frac{1}{2} * frac{1}{2}$?
probability probability-theory
edited Nov 15 at 22:05
David G. Stork
9,11821232
asked Nov 15 at 21:40
Jersey Fonseca
356
So, I know that the general formula to find the probability of independent Bernoulli events is:
$C(n,k)p^kq^{n-k}$
But there are a few questions I have about the application of this theorem to this problem. First off, even though we have 10 indices, we are only concerned with 3 of the options. So I don't know how this formula would apply. Also, I am not sure how to account for the or in this scenario. Is it just that the bit beginning with a 1 is just going to be $frac{1}{2}$ and the one for the 00 a the end is just going to be $frac{1}{2} * frac{1}{2}$?
probability probability-theory
probability probability-theory
edited Nov 15 at 22:05
David G. Stork
9,11821232
asked Nov 15 at 21:40
Jersey Fonseca
356
edited Nov 15 at 22:05
David G. Stork
9,11821232
asked Nov 15 at 21:40
Jersey Fonseca
356
edited Nov 15 at 22:05
David G. Stork
9,11821232
edited Nov 15 at 22:05
David G. Stork
9,11821232
edited Nov 15 at 22:05
David G. Stork
9,11821232
9,11821232
asked Nov 15 at 21:40
Jersey Fonseca
356
asked Nov 15 at 21:40
Jersey Fonseca
356
asked Nov 15 at 21:40
Jersey Fonseca
356
356
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You don't really need to use the Binomial Distribution here. As you noticed in your question,
$$P(text{begins with 1})=frac{1}{2}$$
and
$$P(text{ends with 00})=frac{1}{4}$$
Then use the fact that
$$P(Aspacetext{or}space B)=P(A)+P(B)-P(Aspacetext{and}space B)$$
for any two events $A,B$. If you let $A=text{begins with 1}$ and $B=text{ends with 00}$, you can calculate $P(Aspacetext{or}space B)$, the value you seek, by first calculating $P(Aspacetext{and}space B)$ (which should be a piece of cake, since $A$ and $B$ are independent events).
answered Nov 15 at 21:50
Frpzzd
20.1k638103
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You don't really need to use the Binomial Distribution here. As you noticed in your question,
$$P(text{begins with 1})=frac{1}{2}$$
and
$$P(text{ends with 00})=frac{1}{4}$$
Then use the fact that
$$P(Aspacetext{or}space B)=P(A)+P(B)-P(Aspacetext{and}space B)$$
for any two events $A,B$. If you let $A=text{begins with 1}$ and $B=text{ends with 00}$, you can calculate $P(Aspacetext{or}space B)$, the value you seek, by first calculating $P(Aspacetext{and}space B)$ (which should be a piece of cake, since $A$ and $B$ are independent events).
answered Nov 15 at 21:50
Frpzzd
20.1k638103
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
add a comment |
up vote
1
down vote
accepted
You don't really need to use the Binomial Distribution here. As you noticed in your question,
$$P(text{begins with 1})=frac{1}{2}$$
and
$$P(text{ends with 00})=frac{1}{4}$$
Then use the fact that
$$P(Aspacetext{or}space B)=P(A)+P(B)-P(Aspacetext{and}space B)$$
for any two events $A,B$. If you let $A=text{begins with 1}$ and $B=text{ends with 00}$, you can calculate $P(Aspacetext{or}space B)$, the value you seek, by first calculating $P(Aspacetext{and}space B)$ (which should be a piece of cake, since $A$ and $B$ are independent events).
answered Nov 15 at 21:50
Frpzzd
20.1k638103
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You don't really need to use the Binomial Distribution here. As you noticed in your question,
$$P(text{begins with 1})=frac{1}{2}$$
and
$$P(text{ends with 00})=frac{1}{4}$$
Then use the fact that
$$P(Aspacetext{or}space B)=P(A)+P(B)-P(Aspacetext{and}space B)$$
for any two events $A,B$. If you let $A=text{begins with 1}$ and $B=text{ends with 00}$, you can calculate $P(Aspacetext{or}space B)$, the value you seek, by first calculating $P(Aspacetext{and}space B)$ (which should be a piece of cake, since $A$ and $B$ are independent events).
answered Nov 15 at 21:50
Frpzzd
20.1k638103
You don't really need to use the Binomial Distribution here. As you noticed in your question,
$$P(text{begins with 1})=frac{1}{2}$$
and
$$P(text{ends with 00})=frac{1}{4}$$
Then use the fact that
$$P(Aspacetext{or}space B)=P(A)+P(B)-P(Aspacetext{and}space B)$$
for any two events $A,B$. If you let $A=text{begins with 1}$ and $B=text{ends with 00}$, you can calculate $P(Aspacetext{or}space B)$, the value you seek, by first calculating $P(Aspacetext{and}space B)$ (which should be a piece of cake, since $A$ and $B$ are independent events).
answered Nov 15 at 21:50
Frpzzd
20.1k638103
answered Nov 15 at 21:50
Frpzzd
20.1k638103
answered Nov 15 at 21:50
Frpzzd
20.1k638103
answered Nov 15 at 21:50
Frpzzd
20.1k638103
20.1k638103
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
add a comment |
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
Thanks so much! I appreciate it. That last line was the one I was really missing :)
– Jersey Fonseca
Nov 15 at 21:59
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
@JerseyFonseca No problem! If it helped, don't forget to $checkmark$ and $uparrow$! :D
– Frpzzd
Nov 15 at 22:05
add a comment |
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