Probability Theory - $0$ doesn't appear among $k$ digits chosen randomly?
$begingroup$
What’s the probability that $0$ doesn’t appear among $k$ digits
chosen independently and uniformly at random?
My thinking is that the sample space should be something, say $x$, to the power of $k$ and the event that $0$ doesn't appear is $x-1$ to the power $k$. I am not sure if this is the correct way to think about this. And what the '$x$' should be here.
I saw a solution which says it should be $9^k / 10^k$. But I did not understand how this is.
It would be great if I could get an explanation for this.
probability probability-theory
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
$endgroup$
add a comment |
$begingroup$
What’s the probability that $0$ doesn’t appear among $k$ digits
chosen independently and uniformly at random?
My thinking is that the sample space should be something, say $x$, to the power of $k$ and the event that $0$ doesn't appear is $x-1$ to the power $k$. I am not sure if this is the correct way to think about this. And what the '$x$' should be here.
I saw a solution which says it should be $9^k / 10^k$. But I did not understand how this is.
It would be great if I could get an explanation for this.
probability probability-theory
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
$endgroup$
$begingroup$
There are $9$ non-zero digits. The number of ways of selecting $k$ non-zero digits is then $9times 9times cdots times 9 = 9^k$, seen by rule of product. Similarly, there are $10$ digits and the number of ways of selecting $k$ digits is $10times 10times cdots times 10=10^k$. Noting that each selection is equally likely to occur we have that taking the ratio of these gives the probability.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:14
add a comment |
$begingroup$
What’s the probability that $0$ doesn’t appear among $k$ digits
chosen independently and uniformly at random?
My thinking is that the sample space should be something, say $x$, to the power of $k$ and the event that $0$ doesn't appear is $x-1$ to the power $k$. I am not sure if this is the correct way to think about this. And what the '$x$' should be here.
I saw a solution which says it should be $9^k / 10^k$. But I did not understand how this is.
It would be great if I could get an explanation for this.
probability probability-theory
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
$endgroup$
What’s the probability that $0$ doesn’t appear among $k$ digits
chosen independently and uniformly at random?
My thinking is that the sample space should be something, say $x$, to the power of $k$ and the event that $0$ doesn't appear is $x-1$ to the power $k$. I am not sure if this is the correct way to think about this. And what the '$x$' should be here.
I saw a solution which says it should be $9^k / 10^k$. But I did not understand how this is.
It would be great if I could get an explanation for this.
probability probability-theory
probability probability-theory
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
edited Dec 15 '18 at 3:41
Tianlalu
3,08121038
3,08121038
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
asked Dec 15 '18 at 3:11
jjPotterjjPotter
32
32
$begingroup$
There are $9$ non-zero digits. The number of ways of selecting $k$ non-zero digits is then $9times 9times cdots times 9 = 9^k$, seen by rule of product. Similarly, there are $10$ digits and the number of ways of selecting $k$ digits is $10times 10times cdots times 10=10^k$. Noting that each selection is equally likely to occur we have that taking the ratio of these gives the probability.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:14
add a comment |
$begingroup$
There are $9$ non-zero digits. The number of ways of selecting $k$ non-zero digits is then $9times 9times cdots times 9 = 9^k$, seen by rule of product. Similarly, there are $10$ digits and the number of ways of selecting $k$ digits is $10times 10times cdots times 10=10^k$. Noting that each selection is equally likely to occur we have that taking the ratio of these gives the probability.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:14
$begingroup$
There are $9$ non-zero digits. The number of ways of selecting $k$ non-zero digits is then $9times 9times cdots times 9 = 9^k$, seen by rule of product. Similarly, there are $10$ digits and the number of ways of selecting $k$ digits is $10times 10times cdots times 10=10^k$. Noting that each selection is equally likely to occur we have that taking the ratio of these gives the probability.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:14
$begingroup$
There are $9$ non-zero digits. The number of ways of selecting $k$ non-zero digits is then $9times 9times cdots times 9 = 9^k$, seen by rule of product. Similarly, there are $10$ digits and the number of ways of selecting $k$ digits is $10times 10times cdots times 10=10^k$. Noting that each selection is equally likely to occur we have that taking the ratio of these gives the probability.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:14
add a comment |
1 Answer
1
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$begingroup$
Assuming we're working in base ten, the digits in question are $0, 1, 2, ..., 9$.
Suppose we take a group of $k$ digits, $d_1,d_2,d_3,d_4,...,d_k$.
What is the probability that $d_1 neq 0$? There are $10$ possible digits, and only $1$ is $0$. Thus, $P(d_1 neq 0) = 9/10$.
Similarly, you can say the same for $d_2, d_3, d_4$, and so on: the rationale is the same.
The probability any $d_i$ having a certain value is independent from that of any other $d_j, ineq j$ as well. Thus, the probability of each being nonzero is multiplied together, since they're independent. Thus, for example,
$$P(d_1,d_2 neq 0) = 9/10 cdot 9/10 = (9/10)^2$$
$$P(d_1,d_2,d_3,d_4 neq 0) = 9/10 cdot 9/10 cdot 9/10 cdot 9/10 = (9/10)^4$$
The pattern should then be clear:
$$P(d_1, d_2, ..., d_k neq 0) = (9/10)^k = 9^k/10^k$$
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Assuming we're working in base ten, the digits in question are $0, 1, 2, ..., 9$.
Suppose we take a group of $k$ digits, $d_1,d_2,d_3,d_4,...,d_k$.
What is the probability that $d_1 neq 0$? There are $10$ possible digits, and only $1$ is $0$. Thus, $P(d_1 neq 0) = 9/10$.
Similarly, you can say the same for $d_2, d_3, d_4$, and so on: the rationale is the same.
The probability any $d_i$ having a certain value is independent from that of any other $d_j, ineq j$ as well. Thus, the probability of each being nonzero is multiplied together, since they're independent. Thus, for example,
$$P(d_1,d_2 neq 0) = 9/10 cdot 9/10 = (9/10)^2$$
$$P(d_1,d_2,d_3,d_4 neq 0) = 9/10 cdot 9/10 cdot 9/10 cdot 9/10 = (9/10)^4$$
The pattern should then be clear:
$$P(d_1, d_2, ..., d_k neq 0) = (9/10)^k = 9^k/10^k$$
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
add a comment |
$begingroup$
Assuming we're working in base ten, the digits in question are $0, 1, 2, ..., 9$.
Suppose we take a group of $k$ digits, $d_1,d_2,d_3,d_4,...,d_k$.
What is the probability that $d_1 neq 0$? There are $10$ possible digits, and only $1$ is $0$. Thus, $P(d_1 neq 0) = 9/10$.
Similarly, you can say the same for $d_2, d_3, d_4$, and so on: the rationale is the same.
The probability any $d_i$ having a certain value is independent from that of any other $d_j, ineq j$ as well. Thus, the probability of each being nonzero is multiplied together, since they're independent. Thus, for example,
$$P(d_1,d_2 neq 0) = 9/10 cdot 9/10 = (9/10)^2$$
$$P(d_1,d_2,d_3,d_4 neq 0) = 9/10 cdot 9/10 cdot 9/10 cdot 9/10 = (9/10)^4$$
The pattern should then be clear:
$$P(d_1, d_2, ..., d_k neq 0) = (9/10)^k = 9^k/10^k$$
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
add a comment |
$begingroup$
Assuming we're working in base ten, the digits in question are $0, 1, 2, ..., 9$.
Suppose we take a group of $k$ digits, $d_1,d_2,d_3,d_4,...,d_k$.
What is the probability that $d_1 neq 0$? There are $10$ possible digits, and only $1$ is $0$. Thus, $P(d_1 neq 0) = 9/10$.
Similarly, you can say the same for $d_2, d_3, d_4$, and so on: the rationale is the same.
The probability any $d_i$ having a certain value is independent from that of any other $d_j, ineq j$ as well. Thus, the probability of each being nonzero is multiplied together, since they're independent. Thus, for example,
$$P(d_1,d_2 neq 0) = 9/10 cdot 9/10 = (9/10)^2$$
$$P(d_1,d_2,d_3,d_4 neq 0) = 9/10 cdot 9/10 cdot 9/10 cdot 9/10 = (9/10)^4$$
The pattern should then be clear:
$$P(d_1, d_2, ..., d_k neq 0) = (9/10)^k = 9^k/10^k$$
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
$endgroup$
Assuming we're working in base ten, the digits in question are $0, 1, 2, ..., 9$.
Suppose we take a group of $k$ digits, $d_1,d_2,d_3,d_4,...,d_k$.
What is the probability that $d_1 neq 0$? There are $10$ possible digits, and only $1$ is $0$. Thus, $P(d_1 neq 0) = 9/10$.
Similarly, you can say the same for $d_2, d_3, d_4$, and so on: the rationale is the same.
The probability any $d_i$ having a certain value is independent from that of any other $d_j, ineq j$ as well. Thus, the probability of each being nonzero is multiplied together, since they're independent. Thus, for example,
$$P(d_1,d_2 neq 0) = 9/10 cdot 9/10 = (9/10)^2$$
$$P(d_1,d_2,d_3,d_4 neq 0) = 9/10 cdot 9/10 cdot 9/10 cdot 9/10 = (9/10)^4$$
The pattern should then be clear:
$$P(d_1, d_2, ..., d_k neq 0) = (9/10)^k = 9^k/10^k$$
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
answered Dec 15 '18 at 3:18
Eevee TrainerEevee Trainer
5,4431936
5,4431936
add a comment |
add a comment |
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$begingroup$
There are $9$ non-zero digits. The number of ways of selecting $k$ non-zero digits is then $9times 9times cdots times 9 = 9^k$, seen by rule of product. Similarly, there are $10$ digits and the number of ways of selecting $k$ digits is $10times 10times cdots times 10=10^k$. Noting that each selection is equally likely to occur we have that taking the ratio of these gives the probability.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:14