Calculating probability that the first success is a multiple of $3$
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A biased coin ($P(text{Heads}) = 0.7$) is flipped. If you get the first tails on the $k$th flip, the probability that k is an integer multiple of $3$ can
be expressed as a reduced fraction $frac{a}b$. How can I calculate this probability.
probability
edited Nov 18 at 20:36
Richard
1009
asked Nov 18 at 20:30
helloworld
477
|
show 3 more comments
up vote
0
down vote
favorite
A biased coin ($P(text{Heads}) = 0.7$) is flipped. If you get the first tails on the $k$th flip, the probability that k is an integer multiple of $3$ can
be expressed as a reduced fraction $frac{a}b$. How can I calculate this probability.
probability
edited Nov 18 at 20:36
Richard
1009
asked Nov 18 at 20:30
helloworld
477
1
I'd do all three at once. That is, I'd let $p_0$ be the probability that it comes on a multiple of $3$. And $p_1$ is the probability that it comes on a turn of the form $3k+1$, and $p_2$ corresponds to $3k+2$. Now look what happens over the first three tosses.
– lulu
Nov 18 at 20:36
The probability that k is a multiple of 3 will then always 1/3. Is that correct?
– helloworld
Nov 18 at 20:51
Why would you think that?
– lulu
Nov 18 at 20:51
There is a $.3$ chance that you win on the very first try. Then there is a $left(.7right)^3times .3=.1029$ chance you win on the second. Thus $p_0>.3+.1029>.4$
– lulu
Nov 18 at 20:54
2
Oh, sorry, I wrote down the wrong one. (I was computing $p_1$). To get $p_0$ the winning paths are $left(.7right)^{3k}times left(.7right)^2times .3$.
– lulu
Nov 18 at 21:04
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A biased coin ($P(text{Heads}) = 0.7$) is flipped. If you get the first tails on the $k$th flip, the probability that k is an integer multiple of $3$ can
be expressed as a reduced fraction $frac{a}b$. How can I calculate this probability.
probability
edited Nov 18 at 20:36
Richard
1009
asked Nov 18 at 20:30
helloworld
477
A biased coin ($P(text{Heads}) = 0.7$) is flipped. If you get the first tails on the $k$th flip, the probability that k is an integer multiple of $3$ can
be expressed as a reduced fraction $frac{a}b$. How can I calculate this probability.
probability
probability
edited Nov 18 at 20:36
Richard
1009
asked Nov 18 at 20:30
helloworld
477
edited Nov 18 at 20:36
Richard
1009
asked Nov 18 at 20:30
helloworld
477
edited Nov 18 at 20:36
Richard
1009
edited Nov 18 at 20:36
Richard
1009
edited Nov 18 at 20:36
Richard
1009
1009
asked Nov 18 at 20:30
helloworld
477
asked Nov 18 at 20:30
helloworld
477
asked Nov 18 at 20:30
helloworld
477
477
1
I'd do all three at once. That is, I'd let $p_0$ be the probability that it comes on a multiple of $3$. And $p_1$ is the probability that it comes on a turn of the form $3k+1$, and $p_2$ corresponds to $3k+2$. Now look what happens over the first three tosses.
– lulu
Nov 18 at 20:36
The probability that k is a multiple of 3 will then always 1/3. Is that correct?
– helloworld
Nov 18 at 20:51
Why would you think that?
– lulu
Nov 18 at 20:51
There is a $.3$ chance that you win on the very first try. Then there is a $left(.7right)^3times .3=.1029$ chance you win on the second. Thus $p_0>.3+.1029>.4$
– lulu
Nov 18 at 20:54
2
Oh, sorry, I wrote down the wrong one. (I was computing $p_1$). To get $p_0$ the winning paths are $left(.7right)^{3k}times left(.7right)^2times .3$.
– lulu
Nov 18 at 21:04
|
show 3 more comments
1
I'd do all three at once. That is, I'd let $p_0$ be the probability that it comes on a multiple of $3$. And $p_1$ is the probability that it comes on a turn of the form $3k+1$, and $p_2$ corresponds to $3k+2$. Now look what happens over the first three tosses.
– lulu
Nov 18 at 20:36
The probability that k is a multiple of 3 will then always 1/3. Is that correct?
– helloworld
Nov 18 at 20:51
Why would you think that?
– lulu
Nov 18 at 20:51
There is a $.3$ chance that you win on the very first try. Then there is a $left(.7right)^3times .3=.1029$ chance you win on the second. Thus $p_0>.3+.1029>.4$
– lulu
Nov 18 at 20:54
2
Oh, sorry, I wrote down the wrong one. (I was computing $p_1$). To get $p_0$ the winning paths are $left(.7right)^{3k}times left(.7right)^2times .3$.
– lulu
Nov 18 at 21:04
1
1
I'd do all three at once. That is, I'd let $p_0$ be the probability that it comes on a multiple of $3$. And $p_1$ is the probability that it comes on a turn of the form $3k+1$, and $p_2$ corresponds to $3k+2$. Now look what happens over the first three tosses.
– lulu
Nov 18 at 20:36
I'd do all three at once. That is, I'd let $p_0$ be the probability that it comes on a multiple of $3$. And $p_1$ is the probability that it comes on a turn of the form $3k+1$, and $p_2$ corresponds to $3k+2$. Now look what happens over the first three tosses.
– lulu
Nov 18 at 20:36
The probability that k is a multiple of 3 will then always 1/3. Is that correct?
– helloworld
Nov 18 at 20:51
The probability that k is a multiple of 3 will then always 1/3. Is that correct?
– helloworld
Nov 18 at 20:51
Why would you think that?
– lulu
Nov 18 at 20:51
Why would you think that?
– lulu
Nov 18 at 20:51
There is a $.3$ chance that you win on the very first try. Then there is a $left(.7right)^3times .3=.1029$ chance you win on the second. Thus $p_0>.3+.1029>.4$
– lulu
Nov 18 at 20:54
There is a $.3$ chance that you win on the very first try. Then there is a $left(.7right)^3times .3=.1029$ chance you win on the second. Thus $p_0>.3+.1029>.4$
– lulu
Nov 18 at 20:54
2
2
Oh, sorry, I wrote down the wrong one. (I was computing $p_1$). To get $p_0$ the winning paths are $left(.7right)^{3k}times left(.7right)^2times .3$.
– lulu
Nov 18 at 21:04
Oh, sorry, I wrote down the wrong one. (I was computing $p_1$). To get $p_0$ the winning paths are $left(.7right)^{3k}times left(.7right)^2times .3$.
– lulu
Nov 18 at 21:04
|
show 3 more comments
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1
I'd do all three at once. That is, I'd let $p_0$ be the probability that it comes on a multiple of $3$. And $p_1$ is the probability that it comes on a turn of the form $3k+1$, and $p_2$ corresponds to $3k+2$. Now look what happens over the first three tosses.
– lulu
Nov 18 at 20:36
The probability that k is a multiple of 3 will then always 1/3. Is that correct?
– helloworld
Nov 18 at 20:51
Why would you think that?
– lulu
Nov 18 at 20:51
There is a $.3$ chance that you win on the very first try. Then there is a $left(.7right)^3times .3=.1029$ chance you win on the second. Thus $p_0>.3+.1029>.4$
– lulu
Nov 18 at 20:54
2
Oh, sorry, I wrote down the wrong one. (I was computing $p_1$). To get $p_0$ the winning paths are $left(.7right)^{3k}times left(.7right)^2times .3$.
– lulu
Nov 18 at 21:04