Probability $1$ item is defective out of a sample of $10$? Question
A lifetime of a brand of LEDS has an exponential density with mean 10000 hours. An LED is considered defective if its lifetime is less than 5000 hours. Assuming that the lifetimes of LEDs are independent, what is the probability that in a random sample of 10 LEDs, no more than 1 is defective?
Attempt
I believe this scenario can be modelled with binomial distribution where we have $n = 10$ trials and we want to find the probability that $0$ or $1$ are defective, ie. no more than $1$ is defective.
P($0$ LEDs are defectives out of sample of $10$ or $1$ LED is defective out of a sample of $10$)
$=$ P($0$ LEDs are defectives out of sample of $10$) + P($1$ LED is defective out of sample of $10$)
Let $D$ be the probability that an LED is defective.
$= {10 \choose 0} D^0 (1 - D)^{10} + {10 \choose 1} D^1 (1 - D)^{9}$
Is this correct so far? How do I find the value of $D$, the probability that an LED is defective? (I haven't come across the exponential density before).
 A: What you have done is correct so far.
The exponential distribution with mean $\mu$ is a distribution on the nonnegative real numbers that satisfies the property that 
$$\Pr(X > \mu t) = e^{-t}$$
for all $t \ge 0$. (The notation given on the Wikipedia page is a bit different, but equivalent.)
In particular, for your case, with a mean of $10000$ hours, and as you care about the lamp's lifetime being greater than $5000$ hours (or not), you have, with $L$ denoting the lifetime of a lamp,
$$\Pr\left(L > \frac12 (10000\text{ hours})\right) = e^{-1/2}$$
This is the probability $1-D$ that a lamp is not defective; the probability that it is defective is therefore $D = 1 - e^{-1/2}$.
Plugging that into your calculations, the probability that either all the lamps are non-defective, or that exactly one lamp is defective is
$$(1-D)^{10} + 10 D (1-D)^9 = (e^{-1/2})^{10} + 10(1-e^{-1/2})(e^{-1/2})^9 = e^{-9/2}(10 - 9e^{-1/2}) \approx 0.05.$$
A: The exponential density function is given by
$f(x) = \lambda e^{-\lambda x}$
for positive $x$ and $\mu = \frac{1}{\lambda}$.
If you want to find the probability that one of your LEDs is defective, we would calculate
$P(\mathrm{defective})=\displaystyle\int_0^{5000} \frac{e^{x/10000}}{10000} 
\mathrm{d}x$
just like any other density function.
