Mixing
Poisson distribution (solution check)
Clash Royale CLAN TAG#URR8PPP
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0;
up vote
1
down vote
favorite
A car hire firm has three cars, which it hires out on a daily basis. Number of cars demanded per day follows a poisson distribution with mean 2.1
a) Find the probability that all cars are in use on any one day.
b) Find the probability that all cars are in use on exactly three days of a five day week.
What I tried:
For the a) part: normal poisson formula with mean = 2.1 and x = 3
$$frac(e-2.1)(2.1^3)6$$
Got the answer as 0.1890. The book's answer is 0.350
For the b) part: did poisson to binomial. Found out p = 0.7. Then used 5C3. But I think my method here is wrong.
probability self-study poisson-distribution
edited 2 hours ago
user2974951
419111
asked 4 hours ago
user585380
436
add a comment |Â
up vote
1
down vote
favorite
A car hire firm has three cars, which it hires out on a daily basis. Number of cars demanded per day follows a poisson distribution with mean 2.1
a) Find the probability that all cars are in use on any one day.
b) Find the probability that all cars are in use on exactly three days of a five day week.
What I tried:
For the a) part: normal poisson formula with mean = 2.1 and x = 3
$$frac(e-2.1)(2.1^3)6$$
Got the answer as 0.1890. The book's answer is 0.350
For the b) part: did poisson to binomial. Found out p = 0.7. Then used 5C3. But I think my method here is wrong.
probability self-study poisson-distribution
edited 2 hours ago
user2974951
419111
asked 4 hours ago
user585380
436
Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.)
– Stephan Kolassa
4 hours ago
@StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out.
– user585380
4 hours ago
1
It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution.
– user2974951
4 hours ago
2
@user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve.
– gunes
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A car hire firm has three cars, which it hires out on a daily basis. Number of cars demanded per day follows a poisson distribution with mean 2.1
a) Find the probability that all cars are in use on any one day.
b) Find the probability that all cars are in use on exactly three days of a five day week.
What I tried:
For the a) part: normal poisson formula with mean = 2.1 and x = 3
$$frac(e-2.1)(2.1^3)6$$
Got the answer as 0.1890. The book's answer is 0.350
For the b) part: did poisson to binomial. Found out p = 0.7. Then used 5C3. But I think my method here is wrong.
probability self-study poisson-distribution
edited 2 hours ago
user2974951
419111
asked 4 hours ago
user585380
436
A car hire firm has three cars, which it hires out on a daily basis. Number of cars demanded per day follows a poisson distribution with mean 2.1
a) Find the probability that all cars are in use on any one day.
b) Find the probability that all cars are in use on exactly three days of a five day week.
What I tried:
For the a) part: normal poisson formula with mean = 2.1 and x = 3
$$frac(e-2.1)(2.1^3)6$$
Got the answer as 0.1890. The book's answer is 0.350
For the b) part: did poisson to binomial. Found out p = 0.7. Then used 5C3. But I think my method here is wrong.
probability self-study poisson-distribution
probability self-study poisson-distribution
edited 2 hours ago
user2974951
419111
asked 4 hours ago
user585380
436
edited 2 hours ago
user2974951
419111
asked 4 hours ago
user585380
436
edited 2 hours ago
user2974951
419111
edited 2 hours ago
user2974951
419111
edited 2 hours ago
user2974951
419111
419111
asked 4 hours ago
user585380
436
asked 4 hours ago
user585380
436
asked 4 hours ago
user585380
436
436
Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.)
– Stephan Kolassa
4 hours ago
@StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out.
– user585380
4 hours ago
1
It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution.
– user2974951
4 hours ago
2
@user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve.
– gunes
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
3 hours ago
add a comment |Â
Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.)
– Stephan Kolassa
4 hours ago
@StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out.
– user585380
4 hours ago
1
It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution.
– user2974951
4 hours ago
2
@user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve.
– gunes
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
3 hours ago
Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.)
– Stephan Kolassa
4 hours ago
Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.)
– Stephan Kolassa
4 hours ago
@StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out.
– user585380
4 hours ago
@StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out.
– user585380
4 hours ago
1
1
It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution.
– user2974951
4 hours ago
It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution.
– user2974951
4 hours ago
2
2
@user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve.
– gunes
4 hours ago
@user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve.
– gunes
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
3 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let $Xsim Pois(lambda = 2.1)$ be a Poisson random variable with mean $lambda = 2.1$.
a) Let $A$ be the event all cars are hired in one day. Your mistake is only calculating the probability that $X=3$. In reality, if $X geq 3$ then demand exceeds supply. So even though only $3$ cars are available, the probability is
$P(A) = P(X geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)=0.35$
b) If you can assume that the demand each day is independent, then using the binomial distribution is a correct method. There are $n=5$ days, each with probability $p=0.35$ of demand exceeding supply, so the probability that demand exceeds supply on exactly 3 days is given by
$5 choose 3$$ 0.35^3 (1-0.35)^2$
ie, $P(Y=3)$ where $Y sim Bin(5,0.35)$
answered 4 hours ago
Xiaomi
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
1
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
 |Â
show 2 more comments
up vote
1
down vote
Because, in (a), you need to calculate $P(Xgeq 3)$, which is 0.35 nearly. If the number of requests coming is greater than 3, you still give your three cars. You don't have to have three requests exactly. For (b), you continue with classical binomial.
answered 4 hours ago
gunes
9608
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $Xsim Pois(lambda = 2.1)$ be a Poisson random variable with mean $lambda = 2.1$.
a) Let $A$ be the event all cars are hired in one day. Your mistake is only calculating the probability that $X=3$. In reality, if $X geq 3$ then demand exceeds supply. So even though only $3$ cars are available, the probability is
$P(A) = P(X geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)=0.35$
b) If you can assume that the demand each day is independent, then using the binomial distribution is a correct method. There are $n=5$ days, each with probability $p=0.35$ of demand exceeding supply, so the probability that demand exceeds supply on exactly 3 days is given by
$5 choose 3$$ 0.35^3 (1-0.35)^2$
ie, $P(Y=3)$ where $Y sim Bin(5,0.35)$
answered 4 hours ago
Xiaomi
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
1
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
 |Â
show 2 more comments
up vote
2
down vote
accepted
Let $Xsim Pois(lambda = 2.1)$ be a Poisson random variable with mean $lambda = 2.1$.
a) Let $A$ be the event all cars are hired in one day. Your mistake is only calculating the probability that $X=3$. In reality, if $X geq 3$ then demand exceeds supply. So even though only $3$ cars are available, the probability is
$P(A) = P(X geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)=0.35$
b) If you can assume that the demand each day is independent, then using the binomial distribution is a correct method. There are $n=5$ days, each with probability $p=0.35$ of demand exceeding supply, so the probability that demand exceeds supply on exactly 3 days is given by
$5 choose 3$$ 0.35^3 (1-0.35)^2$
ie, $P(Y=3)$ where $Y sim Bin(5,0.35)$
answered 4 hours ago
Xiaomi
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
1
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $Xsim Pois(lambda = 2.1)$ be a Poisson random variable with mean $lambda = 2.1$.
a) Let $A$ be the event all cars are hired in one day. Your mistake is only calculating the probability that $X=3$. In reality, if $X geq 3$ then demand exceeds supply. So even though only $3$ cars are available, the probability is
$P(A) = P(X geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)=0.35$
b) If you can assume that the demand each day is independent, then using the binomial distribution is a correct method. There are $n=5$ days, each with probability $p=0.35$ of demand exceeding supply, so the probability that demand exceeds supply on exactly 3 days is given by
$5 choose 3$$ 0.35^3 (1-0.35)^2$
ie, $P(Y=3)$ where $Y sim Bin(5,0.35)$
answered 4 hours ago
Xiaomi
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $Xsim Pois(lambda = 2.1)$ be a Poisson random variable with mean $lambda = 2.1$.
a) Let $A$ be the event all cars are hired in one day. Your mistake is only calculating the probability that $X=3$. In reality, if $X geq 3$ then demand exceeds supply. So even though only $3$ cars are available, the probability is
$P(A) = P(X geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)=0.35$
b) If you can assume that the demand each day is independent, then using the binomial distribution is a correct method. There are $n=5$ days, each with probability $p=0.35$ of demand exceeding supply, so the probability that demand exceeds supply on exactly 3 days is given by
$5 choose 3$$ 0.35^3 (1-0.35)^2$
ie, $P(Y=3)$ where $Y sim Bin(5,0.35)$
answered 4 hours ago
Xiaomi
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Xiaomi
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Xiaomi
563
answered 4 hours ago
Xiaomi
563
563
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Xiaomi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
1
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
 |Â
show 2 more comments
2
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
1
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
2
2
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming.
– user585380
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
No problem. Good luck with your studies :)
– Xiaomi
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
4 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1times 5 = 10.5$.
– Xiaomi
3 hours ago
1
1
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day)
– Xiaomi
2 hours ago
 |Â
show 2 more comments
up vote
1
down vote
Because, in (a), you need to calculate $P(Xgeq 3)$, which is 0.35 nearly. If the number of requests coming is greater than 3, you still give your three cars. You don't have to have three requests exactly. For (b), you continue with classical binomial.
answered 4 hours ago
gunes
9608
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
add a comment |Â
up vote
1
down vote
Because, in (a), you need to calculate $P(Xgeq 3)$, which is 0.35 nearly. If the number of requests coming is greater than 3, you still give your three cars. You don't have to have three requests exactly. For (b), you continue with classical binomial.
answered 4 hours ago
gunes
9608
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Because, in (a), you need to calculate $P(Xgeq 3)$, which is 0.35 nearly. If the number of requests coming is greater than 3, you still give your three cars. You don't have to have three requests exactly. For (b), you continue with classical binomial.
answered 4 hours ago
gunes
9608
Because, in (a), you need to calculate $P(Xgeq 3)$, which is 0.35 nearly. If the number of requests coming is greater than 3, you still give your three cars. You don't have to have three requests exactly. For (b), you continue with classical binomial.
answered 4 hours ago
gunes
9608
answered 4 hours ago
gunes
9608
answered 4 hours ago
gunes
9608
answered 4 hours ago
gunes
9608
9608
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
add a comment |Â
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
Is this supposed to be calculated from poisson to binomial.
– user585380
4 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f368579%2fpoisson-distribution-solution-check%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.)
– Stephan Kolassa
4 hours ago
@StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out.
– user585380
4 hours ago
1
It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution.
– user2974951
4 hours ago
2
@user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve.
– gunes
4 hours ago
Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5
– user585380
3 hours ago