Mixing
Question on probability with respect to dice throwing
$begingroup$
I encountered this question while solving few practice problems on probability theory. I have solved it in my own way, but am unsure about whether it's the correct way to go about it.
Four fair dice are thrown. One of them shows a six. Find the probability
that only one die shows a six.
My approach:
$$
begin{align*}
Pbig(text{only 1 six}big|text{at least 1 six}big)
&= frac{P(text{only 1 six}, text{ at least one six})}{P(text{at least 1 six})} \
&= frac{P(text{only 1 six})}{P(text{at least 1 six}) }
end{align*}
$$
Now, $P(text{at least 1 six}) = 1 - frac{5^3}{6^3}$
and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5^3}{6^3}$.
So, substituting, I get $frac{500}{671}$ as my answer.
Is my approach correct?
Any insight would be helpful.
probability probability-theory probability-distributions conditional-probability dice
edited Feb 10 at 11:04
jvdhooft
5,67561641
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
$endgroup$
add a comment |
$begingroup$
I encountered this question while solving few practice problems on probability theory. I have solved it in my own way, but am unsure about whether it's the correct way to go about it.
Four fair dice are thrown. One of them shows a six. Find the probability
that only one die shows a six.
My approach:
$$
begin{align*}
Pbig(text{only 1 six}big|text{at least 1 six}big)
&= frac{P(text{only 1 six}, text{ at least one six})}{P(text{at least 1 six})} \
&= frac{P(text{only 1 six})}{P(text{at least 1 six}) }
end{align*}
$$
Now, $P(text{at least 1 six}) = 1 - frac{5^3}{6^3}$
and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5^3}{6^3}$.
So, substituting, I get $frac{500}{671}$ as my answer.
Is my approach correct?
Any insight would be helpful.
probability probability-theory probability-distributions conditional-probability dice
edited Feb 10 at 11:04
jvdhooft
5,67561641
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
$endgroup$
$begingroup$
When it says "one of them shows a six," does that mean (a) one of the four dice is examined, and it proves to have rolled a six, or (b) all four of the dice are examined, and at least one of them happens to have rolled a six? These are different events.
$endgroup$
– Brian Tung
Jan 26 at 7:11
1
$begingroup$
It means (b). All are examined.
$endgroup$
– LumosMaxima
Jan 26 at 7:12
1
$begingroup$
Your approach and final answer are okay then. Not okay is $P(text{at least 1 six}) = 1 - frac{5}{6^3}$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5}{6^3}$. They should be $P(text{at least 1 six}) = 1 - left(frac{5}{6}right)^3$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}left(frac{5}{6}right)^3$. Because your final answer is okay I suspect they are typos.
$endgroup$
– drhab
Jan 26 at 12:18
$begingroup$
Yes, I fixed them now. Thanks.
$endgroup$
– LumosMaxima
Jan 27 at 8:50
add a comment |
$begingroup$
I encountered this question while solving few practice problems on probability theory. I have solved it in my own way, but am unsure about whether it's the correct way to go about it.
Four fair dice are thrown. One of them shows a six. Find the probability
that only one die shows a six.
My approach:
$$
begin{align*}
Pbig(text{only 1 six}big|text{at least 1 six}big)
&= frac{P(text{only 1 six}, text{ at least one six})}{P(text{at least 1 six})} \
&= frac{P(text{only 1 six})}{P(text{at least 1 six}) }
end{align*}
$$
Now, $P(text{at least 1 six}) = 1 - frac{5^3}{6^3}$
and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5^3}{6^3}$.
So, substituting, I get $frac{500}{671}$ as my answer.
Is my approach correct?
Any insight would be helpful.
probability probability-theory probability-distributions conditional-probability dice
edited Feb 10 at 11:04
jvdhooft
5,67561641
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
$endgroup$
I encountered this question while solving few practice problems on probability theory. I have solved it in my own way, but am unsure about whether it's the correct way to go about it.
Four fair dice are thrown. One of them shows a six. Find the probability
that only one die shows a six.
My approach:
$$
begin{align*}
Pbig(text{only 1 six}big|text{at least 1 six}big)
&= frac{P(text{only 1 six}, text{ at least one six})}{P(text{at least 1 six})} \
&= frac{P(text{only 1 six})}{P(text{at least 1 six}) }
end{align*}
$$
Now, $P(text{at least 1 six}) = 1 - frac{5^3}{6^3}$
and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5^3}{6^3}$.
So, substituting, I get $frac{500}{671}$ as my answer.
Is my approach correct?
Any insight would be helpful.
probability probability-theory probability-distributions conditional-probability dice
probability probability-theory probability-distributions conditional-probability dice
edited Feb 10 at 11:04
jvdhooft
5,67561641
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
edited Feb 10 at 11:04
jvdhooft
5,67561641
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
edited Feb 10 at 11:04
jvdhooft
5,67561641
edited Feb 10 at 11:04
jvdhooft
5,67561641
edited Feb 10 at 11:04
jvdhooft
5,67561641
5,67561641
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
asked Jan 26 at 7:04
LumosMaximaLumosMaxima
858
858
$begingroup$
When it says "one of them shows a six," does that mean (a) one of the four dice is examined, and it proves to have rolled a six, or (b) all four of the dice are examined, and at least one of them happens to have rolled a six? These are different events.
$endgroup$
– Brian Tung
Jan 26 at 7:11
1
$begingroup$
It means (b). All are examined.
$endgroup$
– LumosMaxima
Jan 26 at 7:12
1
$begingroup$
Your approach and final answer are okay then. Not okay is $P(text{at least 1 six}) = 1 - frac{5}{6^3}$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5}{6^3}$. They should be $P(text{at least 1 six}) = 1 - left(frac{5}{6}right)^3$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}left(frac{5}{6}right)^3$. Because your final answer is okay I suspect they are typos.
$endgroup$
– drhab
Jan 26 at 12:18
$begingroup$
Yes, I fixed them now. Thanks.
$endgroup$
– LumosMaxima
Jan 27 at 8:50
add a comment |
$begingroup$
When it says "one of them shows a six," does that mean (a) one of the four dice is examined, and it proves to have rolled a six, or (b) all four of the dice are examined, and at least one of them happens to have rolled a six? These are different events.
$endgroup$
– Brian Tung
Jan 26 at 7:11
1
$begingroup$
It means (b). All are examined.
$endgroup$
– LumosMaxima
Jan 26 at 7:12
1
$begingroup$
Your approach and final answer are okay then. Not okay is $P(text{at least 1 six}) = 1 - frac{5}{6^3}$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5}{6^3}$. They should be $P(text{at least 1 six}) = 1 - left(frac{5}{6}right)^3$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}left(frac{5}{6}right)^3$. Because your final answer is okay I suspect they are typos.
$endgroup$
– drhab
Jan 26 at 12:18
$begingroup$
Yes, I fixed them now. Thanks.
$endgroup$
– LumosMaxima
Jan 27 at 8:50
$begingroup$
When it says "one of them shows a six," does that mean (a) one of the four dice is examined, and it proves to have rolled a six, or (b) all four of the dice are examined, and at least one of them happens to have rolled a six? These are different events.
$endgroup$
– Brian Tung
Jan 26 at 7:11
$begingroup$
When it says "one of them shows a six," does that mean (a) one of the four dice is examined, and it proves to have rolled a six, or (b) all four of the dice are examined, and at least one of them happens to have rolled a six? These are different events.
$endgroup$
– Brian Tung
Jan 26 at 7:11
1
1
$begingroup$
It means (b). All are examined.
$endgroup$
– LumosMaxima
Jan 26 at 7:12
$begingroup$
It means (b). All are examined.
$endgroup$
– LumosMaxima
Jan 26 at 7:12
1
1
$begingroup$
Your approach and final answer are okay then. Not okay is $P(text{at least 1 six}) = 1 - frac{5}{6^3}$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5}{6^3}$. They should be $P(text{at least 1 six}) = 1 - left(frac{5}{6}right)^3$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}left(frac{5}{6}right)^3$. Because your final answer is okay I suspect they are typos.
$endgroup$
– drhab
Jan 26 at 12:18
$begingroup$
Your approach and final answer are okay then. Not okay is $P(text{at least 1 six}) = 1 - frac{5}{6^3}$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5}{6^3}$. They should be $P(text{at least 1 six}) = 1 - left(frac{5}{6}right)^3$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}left(frac{5}{6}right)^3$. Because your final answer is okay I suspect they are typos.
$endgroup$
– drhab
Jan 26 at 12:18
$begingroup$
Yes, I fixed them now. Thanks.
$endgroup$
– LumosMaxima
Jan 27 at 8:50
$begingroup$
Yes, I fixed them now. Thanks.
$endgroup$
– LumosMaxima
Jan 27 at 8:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$textbf{Problem}$. Four dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
$textbf{Solution}$. Since we know in advance that one of the four dice shows a six, we need to use conditional probability.
Let $A$ be the event that at least one of the four dice shows a six, and let
$B$ be the event that exactly one of the four dice shows a six. Since the outcomes in $B$ are contained in the event $A$, we have $Bcap A = B$.
To find $P(A)$, let's find $P(A')$, where $A'$ is the event that no die shows a six ($A'$ is the complement of $A$). Since $P(A')=(5/6)^4$ using a counting technique,
$$
P(A)=1-P(A') = boxed{1-(5/6)^4}.
$$
Now, to find $P(B)$, we'll use the binomial probability distribution:
$$
b(x;n,p) =
begin{cases}
binom{n}{x} p^x (1-p)^{n-x} &mbox{ if }x = 0,1,2,ldots, n, \
qquadquad 0 & mbox{ otherwise},
end{cases}
$$
where $n$ is the number of trials and $p$ is the probability of success.
In this setting, $n=4$, $p=1/6$, and $x=1$.
So
$$
P(B) = b(1; 4,1/6)= binom{4}{1}left(frac{1}{6}right)^1left(frac{5}{6}right)^{3} =boxed{frac{125}{324}}.
$$
Putting this together, we have
$$
P(B|A) = frac{P(Bcap A)}{P(A)} = frac{P(B)}{P(A)} = color{green}{boxed{frac{500}{671}}}.
$$
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
1
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
add a comment |
Your Answer
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$begingroup$
$textbf{Problem}$. Four dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
$textbf{Solution}$. Since we know in advance that one of the four dice shows a six, we need to use conditional probability.
Let $A$ be the event that at least one of the four dice shows a six, and let
$B$ be the event that exactly one of the four dice shows a six. Since the outcomes in $B$ are contained in the event $A$, we have $Bcap A = B$.
To find $P(A)$, let's find $P(A')$, where $A'$ is the event that no die shows a six ($A'$ is the complement of $A$). Since $P(A')=(5/6)^4$ using a counting technique,
$$
P(A)=1-P(A') = boxed{1-(5/6)^4}.
$$
Now, to find $P(B)$, we'll use the binomial probability distribution:
$$
b(x;n,p) =
begin{cases}
binom{n}{x} p^x (1-p)^{n-x} &mbox{ if }x = 0,1,2,ldots, n, \
qquadquad 0 & mbox{ otherwise},
end{cases}
$$
where $n$ is the number of trials and $p$ is the probability of success.
In this setting, $n=4$, $p=1/6$, and $x=1$.
So
$$
P(B) = b(1; 4,1/6)= binom{4}{1}left(frac{1}{6}right)^1left(frac{5}{6}right)^{3} =boxed{frac{125}{324}}.
$$
Putting this together, we have
$$
P(B|A) = frac{P(Bcap A)}{P(A)} = frac{P(B)}{P(A)} = color{green}{boxed{frac{500}{671}}}.
$$
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
1
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
add a comment |
$begingroup$
$textbf{Problem}$. Four dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
$textbf{Solution}$. Since we know in advance that one of the four dice shows a six, we need to use conditional probability.
Let $A$ be the event that at least one of the four dice shows a six, and let
$B$ be the event that exactly one of the four dice shows a six. Since the outcomes in $B$ are contained in the event $A$, we have $Bcap A = B$.
To find $P(A)$, let's find $P(A')$, where $A'$ is the event that no die shows a six ($A'$ is the complement of $A$). Since $P(A')=(5/6)^4$ using a counting technique,
$$
P(A)=1-P(A') = boxed{1-(5/6)^4}.
$$
Now, to find $P(B)$, we'll use the binomial probability distribution:
$$
b(x;n,p) =
begin{cases}
binom{n}{x} p^x (1-p)^{n-x} &mbox{ if }x = 0,1,2,ldots, n, \
qquadquad 0 & mbox{ otherwise},
end{cases}
$$
where $n$ is the number of trials and $p$ is the probability of success.
In this setting, $n=4$, $p=1/6$, and $x=1$.
So
$$
P(B) = b(1; 4,1/6)= binom{4}{1}left(frac{1}{6}right)^1left(frac{5}{6}right)^{3} =boxed{frac{125}{324}}.
$$
Putting this together, we have
$$
P(B|A) = frac{P(Bcap A)}{P(A)} = frac{P(B)}{P(A)} = color{green}{boxed{frac{500}{671}}}.
$$
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
1
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
add a comment |
$begingroup$
$textbf{Problem}$. Four dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
$textbf{Solution}$. Since we know in advance that one of the four dice shows a six, we need to use conditional probability.
Let $A$ be the event that at least one of the four dice shows a six, and let
$B$ be the event that exactly one of the four dice shows a six. Since the outcomes in $B$ are contained in the event $A$, we have $Bcap A = B$.
To find $P(A)$, let's find $P(A')$, where $A'$ is the event that no die shows a six ($A'$ is the complement of $A$). Since $P(A')=(5/6)^4$ using a counting technique,
$$
P(A)=1-P(A') = boxed{1-(5/6)^4}.
$$
Now, to find $P(B)$, we'll use the binomial probability distribution:
$$
b(x;n,p) =
begin{cases}
binom{n}{x} p^x (1-p)^{n-x} &mbox{ if }x = 0,1,2,ldots, n, \
qquadquad 0 & mbox{ otherwise},
end{cases}
$$
where $n$ is the number of trials and $p$ is the probability of success.
In this setting, $n=4$, $p=1/6$, and $x=1$.
So
$$
P(B) = b(1; 4,1/6)= binom{4}{1}left(frac{1}{6}right)^1left(frac{5}{6}right)^{3} =boxed{frac{125}{324}}.
$$
Putting this together, we have
$$
P(B|A) = frac{P(Bcap A)}{P(A)} = frac{P(B)}{P(A)} = color{green}{boxed{frac{500}{671}}}.
$$
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
$textbf{Problem}$. Four dice are thrown. One of them shows a six. Find the probability that only one die shows a six.
$textbf{Solution}$. Since we know in advance that one of the four dice shows a six, we need to use conditional probability.
Let $A$ be the event that at least one of the four dice shows a six, and let
$B$ be the event that exactly one of the four dice shows a six. Since the outcomes in $B$ are contained in the event $A$, we have $Bcap A = B$.
To find $P(A)$, let's find $P(A')$, where $A'$ is the event that no die shows a six ($A'$ is the complement of $A$). Since $P(A')=(5/6)^4$ using a counting technique,
$$
P(A)=1-P(A') = boxed{1-(5/6)^4}.
$$
Now, to find $P(B)$, we'll use the binomial probability distribution:
$$
b(x;n,p) =
begin{cases}
binom{n}{x} p^x (1-p)^{n-x} &mbox{ if }x = 0,1,2,ldots, n, \
qquadquad 0 & mbox{ otherwise},
end{cases}
$$
where $n$ is the number of trials and $p$ is the probability of success.
In this setting, $n=4$, $p=1/6$, and $x=1$.
So
$$
P(B) = b(1; 4,1/6)= binom{4}{1}left(frac{1}{6}right)^1left(frac{5}{6}right)^{3} =boxed{frac{125}{324}}.
$$
Putting this together, we have
$$
P(B|A) = frac{P(Bcap A)}{P(A)} = frac{P(B)}{P(A)} = color{green}{boxed{frac{500}{671}}}.
$$
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
answered Jan 26 at 7:49
Mee Seong ImMee Seong Im
2,8151617
2,8151617
1
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
add a comment |
1
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
1
1
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
$begingroup$
Thank you for the help.
$endgroup$
– LumosMaxima
Jan 27 at 8:51
add a comment |
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$begingroup$
When it says "one of them shows a six," does that mean (a) one of the four dice is examined, and it proves to have rolled a six, or (b) all four of the dice are examined, and at least one of them happens to have rolled a six? These are different events.
$endgroup$
– Brian Tung
Jan 26 at 7:11
1
$begingroup$
It means (b). All are examined.
$endgroup$
– LumosMaxima
Jan 26 at 7:12
1
$begingroup$
Your approach and final answer are okay then. Not okay is $P(text{at least 1 six}) = 1 - frac{5}{6^3}$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}$ $frac{5}{6^3}$. They should be $P(text{at least 1 six}) = 1 - left(frac{5}{6}right)^3$ and $P(text{only 1 six}) = ({}_{4}C_1) frac{1}{6}left(frac{5}{6}right)^3$. Because your final answer is okay I suspect they are typos.
$endgroup$
– drhab
Jan 26 at 12:18
$begingroup$
Yes, I fixed them now. Thanks.
$endgroup$
– LumosMaxima
Jan 27 at 8:50