You could try

Order statistic - Wikipedia, the free encyclopedia
But, if you have not heard of order statistics before then you dont need to learn it to solve this problem. You can use the reasoning i put in my earlier reply.

Here is part (a), step by step, To get you started.

**Concept - the distribution function**
The distribution function of a random variable is normally written F(x). It tells you the probability that the random variable X takes a value less than or equal to some particular value (x).

\(\displaystyle F(x) = P(X \leq x)\)

**Define:**
\(\displaystyle X_1 .... X_{10}\) the lifetime of the 10 components

\(\displaystyle F_i(x) = P(X_i \leq x)\) the distribution function for each X

\(\displaystyle T\) the lifetime of the system

\(\displaystyle G(t) = P(T \leq t)\) the distribution function for T

**Information from question**
\(\displaystyle X_i\) are independant and follow the same distribution: \(\displaystyle exp(\lambda)\)

We are told the mean value of each distribution is 100, and you should know (or be able to look up) that the mean value of an exponential distribution is \(\displaystyle 1/\lambda\)

so \(\displaystyle X_i \sim exp(0.01)\)

You should know (or be able to look up) that the distribution function for he exponential distribution is

\(\displaystyle F_i(x) = 1-e^{-\lambda x} = 1-e^{-0.01x}\)

**Question**
You are told that the system fails when 1 component fails. Find G(t).

**Step 1**
\(\displaystyle G(t) = P(T<t)\)

What is the probability of the system fialing before time t? This happens if at least one component has failed by time t.

\(\displaystyle G(t) = P(T<t)\) =P(at least 1 component failes by time t)

\(\displaystyle G(t) = P(T<t) =1-P(no~component~fails~by~time~t)\)

**Step 2**
What is the probability that no component has failed by time t?

P(no failures) = \(\displaystyle P(X_1~alive~and X_2~alive~and~X_3~alive~and...)\)

P(no failures) = \(\displaystyle P(X_1~alive)P(X_2~alive)P(X_3~alive)P(X_4~alive)...\)

P(no failures) = \(\displaystyle P(X_1 > t)P(X_2 > t)P(X_3 > t)P(X_4>t)...\)

P(no failures) = \(\displaystyle \left(1-F_1(t) \right) \left(1-F_2(t) \right) \left(1-F_3(t) \right) \left(1-F_4(t) \right) \left(1-F_5(t)]\right)...\)

P(no failures) = \(\displaystyle [1-F(t)]^{10}\) (all X's follow the same distribution)

P(no failures) = \(\displaystyle [1-F(t)]^{10}\)

P(no failures) = \(\displaystyle [1-1-e^{-0.01t}]^{10}\)

P(no failures) = \(\displaystyle [e^{-0.01t}]^{10} = e^{-0.1t}\)

**Step 3**
Put it all together:

\(\displaystyle G(t) = P(T<t) =1-P(no~failures~fails~by~time~t)\)

\(\displaystyle G(t) = 1-e^{-0.1t}\)

You can recognise this as an exponential(0.1) distribution.

**Part a Answer**
The lifetime of the system follows an exp(0.1) distribution.