Probabilities involving “at least one” success
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Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has a $.25$ probability of traveling to a time before she was born. During the first year of testing, Isabella uses her machine $5$ times.
Assuming that each trip is equally likely to travel before Isabella was born, what is the probability that at least one trip will travel before Isabella was born?
Round your answer to the nearest hundredth.
I get the answer $1-.75^4$
However, the correct answer is $1-.75^5$. How is this the answer? If finding the complement wouldn't we account for one of the times being before she is born?
probability
asked Nov 22 at 18:13
Jinzu
372412
add a comment |
up vote
1
down vote
favorite
Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has a $.25$ probability of traveling to a time before she was born. During the first year of testing, Isabella uses her machine $5$ times.
Assuming that each trip is equally likely to travel before Isabella was born, what is the probability that at least one trip will travel before Isabella was born?
Round your answer to the nearest hundredth.
I get the answer $1-.75^4$
However, the correct answer is $1-.75^5$. How is this the answer? If finding the complement wouldn't we account for one of the times being before she is born?
probability
asked Nov 22 at 18:13
Jinzu
372412
What does the $P$ mean? The answer is $1-.75^5$
– saulspatz
Nov 22 at 18:17
Yes I should have removed it. It means "the probability of". Why not $1-.75^4$ though? How do you account for the probability of her being born?
– Jinzu
Nov 22 at 18:19
Where did you get the $4$ from? It’s $5$ trips...
– b00n heT
Nov 22 at 18:20
@b00nheT I thought since one of the trips is before being born, that you should subtract one from the total amount of trips.
– Jinzu
Nov 22 at 18:22
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has a $.25$ probability of traveling to a time before she was born. During the first year of testing, Isabella uses her machine $5$ times.
Assuming that each trip is equally likely to travel before Isabella was born, what is the probability that at least one trip will travel before Isabella was born?
Round your answer to the nearest hundredth.
I get the answer $1-.75^4$
However, the correct answer is $1-.75^5$. How is this the answer? If finding the complement wouldn't we account for one of the times being before she is born?
probability
asked Nov 22 at 18:13
Jinzu
372412
Isabella built a time travel machine, but she can't control the destination of her trip. Each time she uses the machine she has a $.25$ probability of traveling to a time before she was born. During the first year of testing, Isabella uses her machine $5$ times.
Assuming that each trip is equally likely to travel before Isabella was born, what is the probability that at least one trip will travel before Isabella was born?
Round your answer to the nearest hundredth.
I get the answer $1-.75^4$
However, the correct answer is $1-.75^5$. How is this the answer? If finding the complement wouldn't we account for one of the times being before she is born?
probability
probability
asked Nov 22 at 18:13
Jinzu
372412
asked Nov 22 at 18:13
Jinzu
372412
edited Nov 22 at 18:18
asked Nov 22 at 18:13
Jinzu
372412
asked Nov 22 at 18:13
Jinzu
372412
asked Nov 22 at 18:13
Jinzu
372412
372412
What does the $P$ mean? The answer is $1-.75^5$
– saulspatz
Nov 22 at 18:17
Yes I should have removed it. It means "the probability of". Why not $1-.75^4$ though? How do you account for the probability of her being born?
– Jinzu
Nov 22 at 18:19
Where did you get the $4$ from? It’s $5$ trips...
– b00n heT
Nov 22 at 18:20
@b00nheT I thought since one of the trips is before being born, that you should subtract one from the total amount of trips.
– Jinzu
Nov 22 at 18:22
add a comment |
What does the $P$ mean? The answer is $1-.75^5$
– saulspatz
Nov 22 at 18:17
Yes I should have removed it. It means "the probability of". Why not $1-.75^4$ though? How do you account for the probability of her being born?
– Jinzu
Nov 22 at 18:19
Where did you get the $4$ from? It’s $5$ trips...
– b00n heT
Nov 22 at 18:20
@b00nheT I thought since one of the trips is before being born, that you should subtract one from the total amount of trips.
– Jinzu
Nov 22 at 18:22
What does the $P$ mean? The answer is $1-.75^5$
– saulspatz
Nov 22 at 18:17
What does the $P$ mean? The answer is $1-.75^5$
– saulspatz
Nov 22 at 18:17
Yes I should have removed it. It means "the probability of". Why not $1-.75^4$ though? How do you account for the probability of her being born?
– Jinzu
Nov 22 at 18:19
Yes I should have removed it. It means "the probability of". Why not $1-.75^4$ though? How do you account for the probability of her being born?
– Jinzu
Nov 22 at 18:19
Where did you get the $4$ from? It’s $5$ trips...
– b00n heT
Nov 22 at 18:20
Where did you get the $4$ from? It’s $5$ trips...
– b00n heT
Nov 22 at 18:20
@b00nheT I thought since one of the trips is before being born, that you should subtract one from the total amount of trips.
– Jinzu
Nov 22 at 18:22
@b00nheT I thought since one of the trips is before being born, that you should subtract one from the total amount of trips.
– Jinzu
Nov 22 at 18:22
add a comment |
1 Answer
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0
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We do not need to account for probability that she was born, it is known to us.
You want at least $1$ case, so why not calculate none, or $0$ case, and subtract it from total number of cases?
Or simply, $$P(at leasthspace{0.2cm} 1)=1 - P(none) $$
$P(none)$ means calculate probability that she always goes in future.
$P(future)=0.75$ for $1$ trip.
So, for $5$ trips, it will be $0.75^5$.
Subtract this from $1$:
You get $$1-0.75^5$$
If this confuses you, calculate discretely for cases: exactly once, exactly twice,thrice, four times and five times.
$$P(exactly hspace{0.2cm} once)=(0.25)cdot (0.75)^4$$
$$P(exactly hspace{0.2cm} twice)=(0.25)^2cdot (0.75)^3$$
$$P(exactly hspace{0.2cm} thrice)=(0.25)^3cdot (0.75)^2$$
$$P(exactly hspace{0.2cm} 4 hspace{0.2cm} times)=(0.25)^4cdot (0.75)^1$$
$$P(exactly hspace{0.2cm} 5 hspace{0.2cm} times)=(0.25)^5cdot (0.75)^0$$
Just add all probabilities: $$(0.25)^1cdot (0.75)^4+(0.25)^2cdot (0.75)^3+(0.25)^3cdot (0.75)^2+(0.25)^4cdot (0.75)^1+(0.25)^5cdot (0.75)^0$$
Verify if you get same result!
answered Nov 22 at 18:27
idea
2,00331024
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
up vote
0
down vote
accepted
We do not need to account for probability that she was born, it is known to us.
You want at least $1$ case, so why not calculate none, or $0$ case, and subtract it from total number of cases?
Or simply, $$P(at leasthspace{0.2cm} 1)=1 - P(none) $$
$P(none)$ means calculate probability that she always goes in future.
$P(future)=0.75$ for $1$ trip.
So, for $5$ trips, it will be $0.75^5$.
Subtract this from $1$:
You get $$1-0.75^5$$
If this confuses you, calculate discretely for cases: exactly once, exactly twice,thrice, four times and five times.
$$P(exactly hspace{0.2cm} once)=(0.25)cdot (0.75)^4$$
$$P(exactly hspace{0.2cm} twice)=(0.25)^2cdot (0.75)^3$$
$$P(exactly hspace{0.2cm} thrice)=(0.25)^3cdot (0.75)^2$$
$$P(exactly hspace{0.2cm} 4 hspace{0.2cm} times)=(0.25)^4cdot (0.75)^1$$
$$P(exactly hspace{0.2cm} 5 hspace{0.2cm} times)=(0.25)^5cdot (0.75)^0$$
Just add all probabilities: $$(0.25)^1cdot (0.75)^4+(0.25)^2cdot (0.75)^3+(0.25)^3cdot (0.75)^2+(0.25)^4cdot (0.75)^1+(0.25)^5cdot (0.75)^0$$
Verify if you get same result!
answered Nov 22 at 18:27
idea
2,00331024
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
add a comment |
up vote
0
down vote
accepted
We do not need to account for probability that she was born, it is known to us.
You want at least $1$ case, so why not calculate none, or $0$ case, and subtract it from total number of cases?
Or simply, $$P(at leasthspace{0.2cm} 1)=1 - P(none) $$
$P(none)$ means calculate probability that she always goes in future.
$P(future)=0.75$ for $1$ trip.
So, for $5$ trips, it will be $0.75^5$.
Subtract this from $1$:
You get $$1-0.75^5$$
If this confuses you, calculate discretely for cases: exactly once, exactly twice,thrice, four times and five times.
$$P(exactly hspace{0.2cm} once)=(0.25)cdot (0.75)^4$$
$$P(exactly hspace{0.2cm} twice)=(0.25)^2cdot (0.75)^3$$
$$P(exactly hspace{0.2cm} thrice)=(0.25)^3cdot (0.75)^2$$
$$P(exactly hspace{0.2cm} 4 hspace{0.2cm} times)=(0.25)^4cdot (0.75)^1$$
$$P(exactly hspace{0.2cm} 5 hspace{0.2cm} times)=(0.25)^5cdot (0.75)^0$$
Just add all probabilities: $$(0.25)^1cdot (0.75)^4+(0.25)^2cdot (0.75)^3+(0.25)^3cdot (0.75)^2+(0.25)^4cdot (0.75)^1+(0.25)^5cdot (0.75)^0$$
Verify if you get same result!
answered Nov 22 at 18:27
idea
2,00331024
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We do not need to account for probability that she was born, it is known to us.
You want at least $1$ case, so why not calculate none, or $0$ case, and subtract it from total number of cases?
Or simply, $$P(at leasthspace{0.2cm} 1)=1 - P(none) $$
$P(none)$ means calculate probability that she always goes in future.
$P(future)=0.75$ for $1$ trip.
So, for $5$ trips, it will be $0.75^5$.
Subtract this from $1$:
You get $$1-0.75^5$$
If this confuses you, calculate discretely for cases: exactly once, exactly twice,thrice, four times and five times.
$$P(exactly hspace{0.2cm} once)=(0.25)cdot (0.75)^4$$
$$P(exactly hspace{0.2cm} twice)=(0.25)^2cdot (0.75)^3$$
$$P(exactly hspace{0.2cm} thrice)=(0.25)^3cdot (0.75)^2$$
$$P(exactly hspace{0.2cm} 4 hspace{0.2cm} times)=(0.25)^4cdot (0.75)^1$$
$$P(exactly hspace{0.2cm} 5 hspace{0.2cm} times)=(0.25)^5cdot (0.75)^0$$
Just add all probabilities: $$(0.25)^1cdot (0.75)^4+(0.25)^2cdot (0.75)^3+(0.25)^3cdot (0.75)^2+(0.25)^4cdot (0.75)^1+(0.25)^5cdot (0.75)^0$$
Verify if you get same result!
answered Nov 22 at 18:27
idea
2,00331024
We do not need to account for probability that she was born, it is known to us.
You want at least $1$ case, so why not calculate none, or $0$ case, and subtract it from total number of cases?
Or simply, $$P(at leasthspace{0.2cm} 1)=1 - P(none) $$
$P(none)$ means calculate probability that she always goes in future.
$P(future)=0.75$ for $1$ trip.
So, for $5$ trips, it will be $0.75^5$.
Subtract this from $1$:
You get $$1-0.75^5$$
If this confuses you, calculate discretely for cases: exactly once, exactly twice,thrice, four times and five times.
$$P(exactly hspace{0.2cm} once)=(0.25)cdot (0.75)^4$$
$$P(exactly hspace{0.2cm} twice)=(0.25)^2cdot (0.75)^3$$
$$P(exactly hspace{0.2cm} thrice)=(0.25)^3cdot (0.75)^2$$
$$P(exactly hspace{0.2cm} 4 hspace{0.2cm} times)=(0.25)^4cdot (0.75)^1$$
$$P(exactly hspace{0.2cm} 5 hspace{0.2cm} times)=(0.25)^5cdot (0.75)^0$$
Just add all probabilities: $$(0.25)^1cdot (0.75)^4+(0.25)^2cdot (0.75)^3+(0.25)^3cdot (0.75)^2+(0.25)^4cdot (0.75)^1+(0.25)^5cdot (0.75)^0$$
Verify if you get same result!
answered Nov 22 at 18:27
idea
2,00331024
edited Nov 22 at 18:35
answered Nov 22 at 18:27
idea
2,00331024
answered Nov 22 at 18:27
idea
2,00331024
answered Nov 22 at 18:27
idea
2,00331024
2,00331024
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
add a comment |
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
@Jinzu have a look at the updated answer.
– idea
Nov 22 at 18:35
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
You are missing some binomial coefficients!
– b00n heT
Nov 22 at 18:45
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
@b00nheT which one? $1$ over 0.25 in exactly once case?
– idea
Nov 22 at 18:49
add a comment |
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What does the $P$ mean? The answer is $1-.75^5$
– saulspatz
Nov 22 at 18:17
Yes I should have removed it. It means "the probability of". Why not $1-.75^4$ though? How do you account for the probability of her being born?
– Jinzu
Nov 22 at 18:19
Where did you get the $4$ from? It’s $5$ trips...
– b00n heT
Nov 22 at 18:20
@b00nheT I thought since one of the trips is before being born, that you should subtract one from the total amount of trips.
– Jinzu
Nov 22 at 18:22