What are the odds to return a product based on its probability to be defective?
$begingroup$
We know that diskettes made by a factory have defects with probability
$0.3$, independently of each other. The factory sells packages containing $10$ diskettes and offers a guarantee that at most one of the $10$ is defective and states that it will refund the purchase money
if the number of defective products exceeds this warranty number. If we buy $3$ packages, what is the probability that one of them will be returned?
MY ATTEMPT
Let us consider a package. If we denote by $X$ the random variable which counts the number of defective diskettes in a package, we conclude that $Xsim B(10,0.3)$. According to this notation, the probability that at most one diskette is defective in the package is given by
begin{align*}
textbf{P}(Xleq 1) = textbf{P}(X = 0) + textbf{P}(X = 1) = {10choose 0}left(frac{3}{10}right)^{0}left(frac{7}{10}right)^{10} + {10choose 1}left(frac{3}{10}right)^{1}left(frac{7}{10}right)^{9}
end{align*}
For convenience, we set $p = textbf{P}(Xleq 1)$. Therefore the probability that exactly one of the three packages will be returned is given by $3p^{2}(1-p)$.
Am I on the right track? I would like to know somebody else's opinion on the subject. Thanks!
probability probability-theory proof-verification probability-distributions
$endgroup$
add a comment |
$begingroup$
We know that diskettes made by a factory have defects with probability
$0.3$, independently of each other. The factory sells packages containing $10$ diskettes and offers a guarantee that at most one of the $10$ is defective and states that it will refund the purchase money
if the number of defective products exceeds this warranty number. If we buy $3$ packages, what is the probability that one of them will be returned?
MY ATTEMPT
Let us consider a package. If we denote by $X$ the random variable which counts the number of defective diskettes in a package, we conclude that $Xsim B(10,0.3)$. According to this notation, the probability that at most one diskette is defective in the package is given by
begin{align*}
textbf{P}(Xleq 1) = textbf{P}(X = 0) + textbf{P}(X = 1) = {10choose 0}left(frac{3}{10}right)^{0}left(frac{7}{10}right)^{10} + {10choose 1}left(frac{3}{10}right)^{1}left(frac{7}{10}right)^{9}
end{align*}
For convenience, we set $p = textbf{P}(Xleq 1)$. Therefore the probability that exactly one of the three packages will be returned is given by $3p^{2}(1-p)$.
Am I on the right track? I would like to know somebody else's opinion on the subject. Thanks!
probability probability-theory proof-verification probability-distributions
$endgroup$
1
$begingroup$
This is the correct answer
$endgroup$
– Peter Foreman
Jan 20 at 18:07
$begingroup$
Thanks for the feedback :)
$endgroup$
– user1337
Jan 20 at 18:08
add a comment |
$begingroup$
We know that diskettes made by a factory have defects with probability
$0.3$, independently of each other. The factory sells packages containing $10$ diskettes and offers a guarantee that at most one of the $10$ is defective and states that it will refund the purchase money
if the number of defective products exceeds this warranty number. If we buy $3$ packages, what is the probability that one of them will be returned?
MY ATTEMPT
Let us consider a package. If we denote by $X$ the random variable which counts the number of defective diskettes in a package, we conclude that $Xsim B(10,0.3)$. According to this notation, the probability that at most one diskette is defective in the package is given by
begin{align*}
textbf{P}(Xleq 1) = textbf{P}(X = 0) + textbf{P}(X = 1) = {10choose 0}left(frac{3}{10}right)^{0}left(frac{7}{10}right)^{10} + {10choose 1}left(frac{3}{10}right)^{1}left(frac{7}{10}right)^{9}
end{align*}
For convenience, we set $p = textbf{P}(Xleq 1)$. Therefore the probability that exactly one of the three packages will be returned is given by $3p^{2}(1-p)$.
Am I on the right track? I would like to know somebody else's opinion on the subject. Thanks!
probability probability-theory proof-verification probability-distributions
$endgroup$
We know that diskettes made by a factory have defects with probability
$0.3$, independently of each other. The factory sells packages containing $10$ diskettes and offers a guarantee that at most one of the $10$ is defective and states that it will refund the purchase money
if the number of defective products exceeds this warranty number. If we buy $3$ packages, what is the probability that one of them will be returned?
MY ATTEMPT
Let us consider a package. If we denote by $X$ the random variable which counts the number of defective diskettes in a package, we conclude that $Xsim B(10,0.3)$. According to this notation, the probability that at most one diskette is defective in the package is given by
begin{align*}
textbf{P}(Xleq 1) = textbf{P}(X = 0) + textbf{P}(X = 1) = {10choose 0}left(frac{3}{10}right)^{0}left(frac{7}{10}right)^{10} + {10choose 1}left(frac{3}{10}right)^{1}left(frac{7}{10}right)^{9}
end{align*}
For convenience, we set $p = textbf{P}(Xleq 1)$. Therefore the probability that exactly one of the three packages will be returned is given by $3p^{2}(1-p)$.
Am I on the right track? I would like to know somebody else's opinion on the subject. Thanks!
probability probability-theory proof-verification probability-distributions
probability probability-theory proof-verification probability-distributions
edited Jan 20 at 18:20
user1337
asked Jan 20 at 18:02
user1337user1337
46110
46110
1
$begingroup$
This is the correct answer
$endgroup$
– Peter Foreman
Jan 20 at 18:07
$begingroup$
Thanks for the feedback :)
$endgroup$
– user1337
Jan 20 at 18:08
add a comment |
1
$begingroup$
This is the correct answer
$endgroup$
– Peter Foreman
Jan 20 at 18:07
$begingroup$
Thanks for the feedback :)
$endgroup$
– user1337
Jan 20 at 18:08
1
1
$begingroup$
This is the correct answer
$endgroup$
– Peter Foreman
Jan 20 at 18:07
$begingroup$
This is the correct answer
$endgroup$
– Peter Foreman
Jan 20 at 18:07
$begingroup$
Thanks for the feedback :)
$endgroup$
– user1337
Jan 20 at 18:08
$begingroup$
Thanks for the feedback :)
$endgroup$
– user1337
Jan 20 at 18:08
add a comment |
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$begingroup$
This is the correct answer
$endgroup$
– Peter Foreman
Jan 20 at 18:07
$begingroup$
Thanks for the feedback :)
$endgroup$
– user1337
Jan 20 at 18:08