Independent Events

The idea of conditional probability is that knowledge of one event can change our understanding of the probability of another. But it's also important to understand when this is not the case.

Events A, B are independent if \Pr(A \mid B) = \Pr(A).

The equation \Pr(A\mid B) = \Pr(A) very concretely says: knowledge that the event B has occurred does not change our understanding of \Pr(A). Assuming \Pr(A) \neq 0, this is equivalent to \Pr(A\cap B) = \Pr(A)\Pr(B), since: \Pr(A\mid B) \amp = \Pr(A) \frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A) \Pr(A\cap B) \amp = \Pr(A)\Pr(B) This latter form of the equation is often more useful in calculations.

In , we considered an experiment in which we rolled a fair 6-sided die twice. We defined events A that the sum of the rolls is at least 10 and B that the first roll is a 6, and we found that \Pr(A \mid B) \neq \Pr(A), so A and B are not independent.

Now consider the event C that the sum of the rolls is 7, which sounds very similar to the event A. We have: C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \Pr(C) \amp = \frac{6}{36} = \frac{1}{6} B\cap C \amp = \{(6, 1)\} \Pr(B\cap C) \amp = \frac{1}{36} \text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C) So, even though the descriptions of events A and C are very similar, the event C is independent with B, while A is not.

An experiment consists of rolling a fair die two times. Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first. Are A and B independent?

A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 4), (3, 5), (3, 6), \amp (4, 5), (4, 6), \amp (5, 6)\} A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), so A and B are not independent.

An experiment consists of flipping a fair coin three times. Let A be the event that the first and second flips match. Let B be the event that there are at least two heads. Are A and B independent?

A \amp = \{ HHH, HHT, TTH, TTT \} B \amp = \{ HHH, HHT, HTH, THH \} A\cap B \amp = \{HHH, HHT\} So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. Finally: \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), so A and B are independent.

Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the sample space \Omega = \{1, 2, 3, 4, 5, 6\}. Create a probability distribution for \Omega so that A, B are independent.

Example Distribution x \Pr(x) 1 0.1 2 0.2 3 0.2 4 0.1 5 0.1 6 0.3

Now \Pr(A) = 0.5, \Pr(B) = 0.4, and \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B), so A and B are independent.