Question

An urn contains four tickets with numbers 112, 121, 211, 222 and one ticket is drawn. Let $A_i(i=1,2,3)$ be the event that the ith digit of the number if tickets drawn is 1. Discuss the independence of the events $A_1,A_2,A_3.$. If they are independent enter 1, else enter 0.

Answer 1

We have $P\left(A_1\right)=\frac{2}{4}=\frac{1}{2}=P\left(A_2\right)=P\left(A_3\right).$

[Note that $P\left(A_1\right)$ is the probability of the event that the first digit is 1 and since there are two numbers having 1 at the place out of four, we have $P\left(A_1\right)=\frac{2}{4}=\frac{1}{2}.$ Similarly for $P\left(A_2\right)$ and $P\left(A_3\right)$].

$A_1\cap A_2$ is the event that the first two digits in the numbers drawn are each equal to 1 and

so $P(A_1\cap A_2)=\frac{1}{4}=\frac{1}{2}.\frac{1}{2}=P\left(A_1\right)P\left(A_2\right).$

Similarly, $P(A_2\cap A_3)=P\left(A_2\right)P\left(A_3\right)$ and $P(A_3\cap A_1)=P\left(A_3\right)P\left(A_1\right).$

Thus the events $A_1,A_2$ and $A_3$ are equal to 1

and since there is no such number, we have $P(A_1\cap A_2\cap A_3)=0\ne P\left(A_1\right)P\left(A_2\right)P\left(A_3\right).$

Hence the events $A_1,A_2,A_3$ are not mutually independent although they are pairwise independent.