# How to solve Geometric distribution problem?

A plane’s engines start successfully at a given attempt with a probability of 0.75. Any time that the mechanics are unsuccessful in starting the engines, they must wait five minutes before trying again.

Find probability that the plane is launched within 10 minutes.

I was given a solution

$$1-0.25^3 = 0.984$$

But what is formula for it?

Pay attention that at most the mechanics can try 3 times. This means that the successful start defines 3 cases.
The probability to start in the first try is : $0.75$
The probability to start in the second try is : $0.75\cdot0.25$
The probability to start in the third try is : $0.75\cdot0.25\cdot0.25$
The problem is that the cases are not disjoint and you can not calculate the requested probability by their sum.
You can use conditional probability but there is an easier approach. The probability to start the plane in 10 minutes equals 1 - the probability not to start the plane in the 10 minutes, meaning you fail in all the first three attempts. The probability to fail in three attempts equals $(1-0.75)^3$ therefore the correct answer is $1-0.25^3$
A geometric distribution gives you the probability that the first success happens in the $k$ try. That means you need to fail $k-1$ times and succeed once. Let $p$ be the probability of success (in your case $p=0.75$) then, the probability that the first success happens in the $k$ try equals: $P(k)=(1-p)^{k-1} \cdot p$.

• it is not correct Commented Mar 4, 2017 at 14:44
• because answer is 1-0.25^3 = 0.984 Commented Mar 4, 2017 at 14:44

So the formula that the solution uses is: $$P(N \geq a) = (1-p)^{a-1}$$ In this instance, we want to find

\begin{align*} P(N \leq 3) &= 1 - P(N \geq 4)\\ &= 1 - (1-.75)^3 \end{align*}