Mixing
Two coins, one experiment: what is the probability that the coin lands on heads on exactly $7$ of the $10$...
$begingroup$
When coin $1$ is flipped, it lands on heads with probability $0.4$; when coin $2$ is flipped, it lands on heads with probability $0.7$. One of these coins is randomly chosen and flipped $10$ times.
(a) What is the probability that the coin lands on heads on
exactly $7$ of the $10$ flips?
(b) Given that the first of these $10$ flips lands heads, what is the conditional probability that exactly $7$ of the $10$ flips land on heads?
MY ATTEMPT
(a) Let us denote by $X$ the random variable which counts the numbers of heads in 10 flips. Thus $Xsimtext{Binomial}(10,p)$, where the parameter $p$ depends on the coin that is chosen: $C = 1$ stands for the event "the first coin is chosen" and $C = 2$ stands for the event "the second coin is chosen". Based on these considerations, we have the answer to the first question, which is
begin{align*}
textbf{P}(X = 7) = textbf{P}(X = 7mid C = 1)textbf{P}(C = 1) + textbf{P}(X = 7mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(X = 7mid C = 1) = displaystyle{10choose 7}(0.4)^{7}(0,6)^{3}\\
textbf{P}(X = 7mid C = 2) = displaystyle{10choose 7}(0.7)^{7}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
(b) Since $X = displaystylesum_{k=1}^{10} X_{k}$, where each $X_{k}simtext{Bernoulli}(p)$, we are interested in the event $textbf{P}(X = 7mid X_{1} = 1)$, which is equivalent to
begin{align*}
textbf{P}(X = 7mid X_{1} = 1) & = frac{textbf{P}(X = 7, X_{1} = 1)}{textbf{P}(X_{1} = 1)} = frac{textbf{P}left(displaystylesum_{k=2}^{10}X_{k} = 6right)}{textbf{P}(X_{1} = 1)}
end{align*}
Since $textbf{P}(X_{1} = 1) = textbf{P}(X_{1} = 1 | C = 1)textbf{P}(C = 1) + textbf{P}(X_{1} = 1 | C = 2)textbf{P}(C = 2)$, it suffices to calculate $textbf{P}(Y = 6)$, where $Y := displaystylesum_{k=2}^{10}X_{k}$ and, consequentely, $Ysimtext{Binomial}(9,p)$:
begin{align*}
textbf{P}(Y = 6) = textbf{P}(Y = 6mid C = 1)textbf{P}(C = 1) + textbf{P}(Y = 6mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(Y = 6mid C = 1) = displaystyle{9choose 6}(0.4)^{6}(0,6)^{3}\\
textbf{P}(Y = 6mid C = 2) = displaystyle{9choose 6}(0.7)^{6}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
My question is: am I working it right or is there any conceptual misapplication?
Any contribution is appreciated. Thanks in advance!
probability probability-theory conditional-probability binomial-distribution
edited Jan 27 at 2:45
Key Flex
8,63561233
asked Jan 27 at 2:21
user1337user1337
46310
$endgroup$
add a comment |
$begingroup$
When coin $1$ is flipped, it lands on heads with probability $0.4$; when coin $2$ is flipped, it lands on heads with probability $0.7$. One of these coins is randomly chosen and flipped $10$ times.
(a) What is the probability that the coin lands on heads on
exactly $7$ of the $10$ flips?
(b) Given that the first of these $10$ flips lands heads, what is the conditional probability that exactly $7$ of the $10$ flips land on heads?
MY ATTEMPT
(a) Let us denote by $X$ the random variable which counts the numbers of heads in 10 flips. Thus $Xsimtext{Binomial}(10,p)$, where the parameter $p$ depends on the coin that is chosen: $C = 1$ stands for the event "the first coin is chosen" and $C = 2$ stands for the event "the second coin is chosen". Based on these considerations, we have the answer to the first question, which is
begin{align*}
textbf{P}(X = 7) = textbf{P}(X = 7mid C = 1)textbf{P}(C = 1) + textbf{P}(X = 7mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(X = 7mid C = 1) = displaystyle{10choose 7}(0.4)^{7}(0,6)^{3}\\
textbf{P}(X = 7mid C = 2) = displaystyle{10choose 7}(0.7)^{7}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
(b) Since $X = displaystylesum_{k=1}^{10} X_{k}$, where each $X_{k}simtext{Bernoulli}(p)$, we are interested in the event $textbf{P}(X = 7mid X_{1} = 1)$, which is equivalent to
begin{align*}
textbf{P}(X = 7mid X_{1} = 1) & = frac{textbf{P}(X = 7, X_{1} = 1)}{textbf{P}(X_{1} = 1)} = frac{textbf{P}left(displaystylesum_{k=2}^{10}X_{k} = 6right)}{textbf{P}(X_{1} = 1)}
end{align*}
Since $textbf{P}(X_{1} = 1) = textbf{P}(X_{1} = 1 | C = 1)textbf{P}(C = 1) + textbf{P}(X_{1} = 1 | C = 2)textbf{P}(C = 2)$, it suffices to calculate $textbf{P}(Y = 6)$, where $Y := displaystylesum_{k=2}^{10}X_{k}$ and, consequentely, $Ysimtext{Binomial}(9,p)$:
begin{align*}
textbf{P}(Y = 6) = textbf{P}(Y = 6mid C = 1)textbf{P}(C = 1) + textbf{P}(Y = 6mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(Y = 6mid C = 1) = displaystyle{9choose 6}(0.4)^{6}(0,6)^{3}\\
textbf{P}(Y = 6mid C = 2) = displaystyle{9choose 6}(0.7)^{6}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
My question is: am I working it right or is there any conceptual misapplication?
Any contribution is appreciated. Thanks in advance!
probability probability-theory conditional-probability binomial-distribution
edited Jan 27 at 2:45
Key Flex
8,63561233
asked Jan 27 at 2:21
user1337user1337
46310
$endgroup$
add a comment |
$begingroup$
When coin $1$ is flipped, it lands on heads with probability $0.4$; when coin $2$ is flipped, it lands on heads with probability $0.7$. One of these coins is randomly chosen and flipped $10$ times.
(a) What is the probability that the coin lands on heads on
exactly $7$ of the $10$ flips?
(b) Given that the first of these $10$ flips lands heads, what is the conditional probability that exactly $7$ of the $10$ flips land on heads?
MY ATTEMPT
(a) Let us denote by $X$ the random variable which counts the numbers of heads in 10 flips. Thus $Xsimtext{Binomial}(10,p)$, where the parameter $p$ depends on the coin that is chosen: $C = 1$ stands for the event "the first coin is chosen" and $C = 2$ stands for the event "the second coin is chosen". Based on these considerations, we have the answer to the first question, which is
begin{align*}
textbf{P}(X = 7) = textbf{P}(X = 7mid C = 1)textbf{P}(C = 1) + textbf{P}(X = 7mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(X = 7mid C = 1) = displaystyle{10choose 7}(0.4)^{7}(0,6)^{3}\\
textbf{P}(X = 7mid C = 2) = displaystyle{10choose 7}(0.7)^{7}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
(b) Since $X = displaystylesum_{k=1}^{10} X_{k}$, where each $X_{k}simtext{Bernoulli}(p)$, we are interested in the event $textbf{P}(X = 7mid X_{1} = 1)$, which is equivalent to
begin{align*}
textbf{P}(X = 7mid X_{1} = 1) & = frac{textbf{P}(X = 7, X_{1} = 1)}{textbf{P}(X_{1} = 1)} = frac{textbf{P}left(displaystylesum_{k=2}^{10}X_{k} = 6right)}{textbf{P}(X_{1} = 1)}
end{align*}
Since $textbf{P}(X_{1} = 1) = textbf{P}(X_{1} = 1 | C = 1)textbf{P}(C = 1) + textbf{P}(X_{1} = 1 | C = 2)textbf{P}(C = 2)$, it suffices to calculate $textbf{P}(Y = 6)$, where $Y := displaystylesum_{k=2}^{10}X_{k}$ and, consequentely, $Ysimtext{Binomial}(9,p)$:
begin{align*}
textbf{P}(Y = 6) = textbf{P}(Y = 6mid C = 1)textbf{P}(C = 1) + textbf{P}(Y = 6mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(Y = 6mid C = 1) = displaystyle{9choose 6}(0.4)^{6}(0,6)^{3}\\
textbf{P}(Y = 6mid C = 2) = displaystyle{9choose 6}(0.7)^{6}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
My question is: am I working it right or is there any conceptual misapplication?
Any contribution is appreciated. Thanks in advance!
probability probability-theory conditional-probability binomial-distribution
edited Jan 27 at 2:45
Key Flex
8,63561233
asked Jan 27 at 2:21
user1337user1337
46310
$endgroup$
When coin $1$ is flipped, it lands on heads with probability $0.4$; when coin $2$ is flipped, it lands on heads with probability $0.7$. One of these coins is randomly chosen and flipped $10$ times.
(a) What is the probability that the coin lands on heads on
exactly $7$ of the $10$ flips?
(b) Given that the first of these $10$ flips lands heads, what is the conditional probability that exactly $7$ of the $10$ flips land on heads?
MY ATTEMPT
(a) Let us denote by $X$ the random variable which counts the numbers of heads in 10 flips. Thus $Xsimtext{Binomial}(10,p)$, where the parameter $p$ depends on the coin that is chosen: $C = 1$ stands for the event "the first coin is chosen" and $C = 2$ stands for the event "the second coin is chosen". Based on these considerations, we have the answer to the first question, which is
begin{align*}
textbf{P}(X = 7) = textbf{P}(X = 7mid C = 1)textbf{P}(C = 1) + textbf{P}(X = 7mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(X = 7mid C = 1) = displaystyle{10choose 7}(0.4)^{7}(0,6)^{3}\\
textbf{P}(X = 7mid C = 2) = displaystyle{10choose 7}(0.7)^{7}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
(b) Since $X = displaystylesum_{k=1}^{10} X_{k}$, where each $X_{k}simtext{Bernoulli}(p)$, we are interested in the event $textbf{P}(X = 7mid X_{1} = 1)$, which is equivalent to
begin{align*}
textbf{P}(X = 7mid X_{1} = 1) & = frac{textbf{P}(X = 7, X_{1} = 1)}{textbf{P}(X_{1} = 1)} = frac{textbf{P}left(displaystylesum_{k=2}^{10}X_{k} = 6right)}{textbf{P}(X_{1} = 1)}
end{align*}
Since $textbf{P}(X_{1} = 1) = textbf{P}(X_{1} = 1 | C = 1)textbf{P}(C = 1) + textbf{P}(X_{1} = 1 | C = 2)textbf{P}(C = 2)$, it suffices to calculate $textbf{P}(Y = 6)$, where $Y := displaystylesum_{k=2}^{10}X_{k}$ and, consequentely, $Ysimtext{Binomial}(9,p)$:
begin{align*}
textbf{P}(Y = 6) = textbf{P}(Y = 6mid C = 1)textbf{P}(C = 1) + textbf{P}(Y = 6mid C = 2)textbf{P}(C = 2)
end{align*}
where
begin{cases}
textbf{P}(Y = 6mid C = 1) = displaystyle{9choose 6}(0.4)^{6}(0,6)^{3}\\
textbf{P}(Y = 6mid C = 2) = displaystyle{9choose 6}(0.7)^{6}(0,3)^{3}\\
textbf{P}(C = 1) = textbf{P}(C = 2) = 0.5
end{cases}
My question is: am I working it right or is there any conceptual misapplication?
Any contribution is appreciated. Thanks in advance!
probability probability-theory conditional-probability binomial-distribution
probability probability-theory conditional-probability binomial-distribution
edited Jan 27 at 2:45
Key Flex
8,63561233
asked Jan 27 at 2:21
user1337user1337
46310
edited Jan 27 at 2:45
Key Flex
8,63561233
asked Jan 27 at 2:21
user1337user1337
46310
edited Jan 27 at 2:45
Key Flex
8,63561233
edited Jan 27 at 2:45
Key Flex
8,63561233
edited Jan 27 at 2:45
Key Flex
8,63561233
8,63561233
asked Jan 27 at 2:21
user1337user1337
46310
asked Jan 27 at 2:21
user1337user1337
46310
asked Jan 27 at 2:21
user1337user1337
46310
46310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=dfrac{P(Hcap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(Hcap T)=P(Hcap T|K_1)P(K_1)+P(Hcap T|K_2)P(K_2)$
$$=dbinom96(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$$
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
$endgroup$
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
1
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=dfrac{P(Hcap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(Hcap T)=P(Hcap T|K_1)P(K_1)+P(Hcap T|K_2)P(K_2)$
$$=dbinom96(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$$
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
$endgroup$
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
1
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
add a comment |
$begingroup$
For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=dfrac{P(Hcap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(Hcap T)=P(Hcap T|K_1)P(K_1)+P(Hcap T|K_2)P(K_2)$
$$=dbinom96(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$$
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
$endgroup$
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
1
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
add a comment |
$begingroup$
For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=dfrac{P(Hcap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(Hcap T)=P(Hcap T|K_1)P(K_1)+P(Hcap T|K_2)P(K_2)$
$$=dbinom96(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$$
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
$endgroup$
For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=dfrac{P(Hcap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(Hcap T)=P(Hcap T|K_1)P(K_1)+P(Hcap T|K_2)P(K_2)$
$$=dbinom96(0.4)^6(0.6)^3(0.5)+dbinom96(0.7)^6(0.3)^3(0.5)$$
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
edited Jan 27 at 2:43
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
answered Jan 27 at 2:32
Key FlexKey Flex
8,63561233
8,63561233
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
1
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
add a comment |
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
1
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
$begingroup$
In the first place, thanks for the contribution Flex! As far as I have checked, we have given the same answers. Maybe there is just one typo: it should be $(0,4)^{6}(0.6)^{3}(0.5)$ and $(0.7)^{6}(0.3)^{3}(0.5)$ in the last line.
$endgroup$
– user1337
Jan 27 at 2:39
1
1
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
$begingroup$
@user1337 Fixed it! Thanks
$endgroup$
– Key Flex
Jan 27 at 2:44
add a comment |
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