1. We find $\Pr(Y\le y)$. This is $\Pr(a+(b-a)X)\le y$, which is $\Pr\left(X\le \frac{y-a}{b-a}\right)$.
If $y-a\le 0$, that is, if $y\le a$, this probability is $0$.
If $\dfrac{y-a}{b-a}\ge 1$, that is, if $y\ge b$, this probability is $1$.
And finally, the interesting part. If $a\lt y\lt b$, this probability is $\dfrac{y-a}{b-a}$.
The three sentences above describe the cumulative distribution function of $Y$. For the density, differentiate. We get density $\dfrac{1}{b-a}$ on $(a,b)$ and $0$ elsewhere. Note that $Y$ has uniform distribution on $(a,b)$ (or equivalently, $[a,b]$).
2. a) Our distributions are continuous. So, as the problem is currently stated, the probability is $0$.
For b), each of $X_1$ and $X_2$ has density function of the shape $e^{-t/1000}$ when $t\gt 0$. So the probability the first is alive after $1200$ hours, by integration, is $e^{-1200/1000}$. The same is true for the second.
Because $X_1$ and $X_2$ are independent, the probability both components are still alive after $1200$ hours is the product of the individual probabilities, that is, $e^{-2400/1000}$. If we interpret "device fails after $1200$ hours" as meaning that the lifetime of the device is at least $1200$, that gives the answer to the question.
Second answer, for a different interpretation: Batteries cannot die before they go into the flashlight. Because this interpretation involves both the
minimum of two exponentials and the sum of several exponentials, it makes a
more interesting problem than did the assumptions in my first answer. It is the interpretation suggested in Andre's note and used copper.hat's multiple integration.
Step 1: Wait for one of two initial batteries to fail. This waiting time is
the minimum of two exponentials with failure rate $\lambda$, and hence
$X_1$ ~ EXP($2\lambda$).
Step 2: Throw out dead battery, replace with new one. By the no-memory property,
the one of the two batteries in the flashlight that did not die is as good
as new. Waiting time for one of these two batteries to die is again
$X_2$ ~ EXP($2\lambda$).
Last step $n-1$; Throw out dead battery, replace with $n$th (last remaining
replacement) battery: Light goes out after additional time
$X_{n-1}$ ~ EXP($2\lambda$).
Total time flashlight is lit is $T = X_1 + \dots + X_{n-1}$. This is the
sum of $(n-1)$ exponentials, so $T$ ~ GAMMA($n-1,$ $2\lambda$).
This is a gamma distribution with shape parameter $n-1$ and rate parameter $2\lambda.$ When the shape parameter is a positive integer the gamma
distribution is sometimes called an Erlang distribution (especially in
queueing theory).
Check: A previous answer, apparently using the same assumptions and with $n=3,$
has the CDF of the random variable $T$ as
$F_T(x) = 1 - \exp(-2\lambda x)(1 + 2\lambda x),$ for $x > 0$. The form of
the CDF does indeed get messier with increasing $n$, but the mean and variance
are simple expressions in $n$ and $\lambda.$
In R, we easily verify (in one instance, anyhow) that this is a special case of the gamma (Erlang)
distribution. Let $n = 3$, $\lambda = 1/15$, and $x = 1$. So this is the
(small) probability that the flashlight goes dark by time 1.
The code 'pgamma(1, 2, 2/15)' and the code '1 - exp(-2/15)*(1 + 2/15)' both
return the probability 0.008136905. Also,
'qgamma(.5, 2, 2/15)' finds the exact median time the flaslhight burns to be 12.58760, and
'mean(rgamma(10^5, 2, 2/15))' approximates the mean as
14.97 (exact is 15).
Best Answer
Since $y_1 < y_2$
$P(Y_1 \leq y_1 , Y_2 \leq y_2) = P(\min(X_1, X_2) \leq y_1 \text{ and } \max(X_1, X_2) \leq y_2) = P( \{ X_1 \leq y_1 \text{ or } X_2 \leq y_1 \} \text{ and } X_1 \leq y_2 \text{ and } X_2 \leq y_2) = P(\{X_1 \leq y_1 \text{ and } X_2 \leq y_2 \} \text{ or } \{X_1 \leq y_2 \text{ and } X_2 \leq y_1 \})$
Now, just use the fact that $P(A \text{ or } B) = P(A) + P(B) - P(A\text{ and } B)$.