In a new fiber-optic communication system, transmission errors occur at a rate of 1.5 per 20 seconds. What is the probability that the first error will occur between 15 seconds and 45 seconds?
There's also a second part of this problem where instead of the first error it's the fourth error.
How do I solve this problem? Thanks for your time.
$$p=1.5/20 = 0.075 = 7.5\%$$
So there is a $7.5\%$ chance of a transmission error in any given second.
Let the interval $0\leq t\leq 1$ be the $1$st second, $1\leq t\leq 2$ be the $2$nd, etc. The probability that $k-1$ errors will happen in $n-1$ seconds is $$\left(\begin{matrix}n-1\\k-1\end{matrix}\right)(1-p)^{n-k}p^{k-1}$$
Then the probability of the $k$th error happening in $n$th second is $p$ times this value, or $$\left(\begin{matrix}n-1\\k-1\end{matrix}\right)(1-p)^{n-k}p^k$$
So your probability of the first error happening between $15$ and $45$ seconds is $$\sum_{n=16}^{45} \left(\begin{matrix}n-1\\0\end{matrix}\right)(1-p)^{n-1}p=\sum_{n=16}^{45} (0.925)^{n-1}(0.075) \approx 0.2806$$
and the probability of the fourth error happening is $$\sum_{n=16}^{45} \left(\begin{matrix}n-1\\3\end{matrix}\right)(1-p)^{n-4}p^4 =\sum_{n=16}^{45} \left(\begin{matrix}n-1\\3\end{matrix}\right)(0.925)^{n-4}(0.075)^4 \approx 0.4173 $$ I don't include $n=15$ because that represents the probability the $n$th error will occur between $14$ and $15$ seconds.