Question

Let $A_1,A_2,\dots ,A_n$be arbitrary events, and define

$Ck=${at least $k$of the $Ai$occur}. Show that

$\sum _{k=1}^{n}P\left(C_k\right)=\sum _{k=1}^{n}P\left(A_k\right)$

Hint: Let $X$denote the number of the $Ai$that occur. Show

that both sides of the preceding equation are equal to $E\left[X\right]$.

Short answer

The arbitrary events is showed as$\sum _{k=1}^{n}P\left(C_k\right)=E\left(X\right)=\sum _{k=1}^{n}P\left(A_k\right)$.

Step-by-step solution

Given Information

$P\left(C_k\right)=P(X\ge k)$as arbitrary events in$A_1,A_2,\dots ,A_n$.

Explanation

We have that,$P\left(C_k\right)=P(X\ge k)$

Hence,$\sum _{k=1}^{n}P\left(C_k\right)=\sum _{k=1}^{n}P(X\ge k)=\sum _{k=0}^{\infty }P(X>k)=E\left(X\right)$

where the last equality is the famous expression of the mean of non-negative discrete random variable. On the other hand, define random variables $I_i$to be indicators whether event $A_i$has occurred or not.

Explanation

We have that, $X=\sum _{k=1}^n{I}_k$

Because of the linearity of the mean, we have that,

$E\left(X\right)=\sum _{k=1}^{n}E\left(I_k\right)=\sum _{k=1}^{n}P\left(I_k=1\right)=\sum _{k=1}^{n}P\left(A_k\right)$

so we have showed that,$\sum _{k=1}^{n}P\left(C_k\right)=E\left(X\right)=\sum _{k=1}^{n}P\left(A_k\right)$.

Final answer

The arbitrary events is showed as$\sum _{k=1}^{n}P\left(C_k\right)=E\left(X\right)=\sum _{k=1}^{n}P\left(A_k\right)$.