Q. 7.5 - A First Course in Probability

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 $E\left[X\right]$.

Step-by-Step Solution

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 1: Given Information

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

Step 2: 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. 

Step 3: 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, $\sum _{k=1}^{n}P\left(C_k\right)=E\left(X\right)=\sum _{k=1}^{n}P\left(A_k\right)$.

Step 4: Final answer

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